6. How much heat energy is required to change a 0.3 kg ice cube from a solid at -20 °C to steam at 120 °℃ ?​

Answers

Answer 1

The heat energy required to change a 0.3 kg ice cube from a solid at -20 °C to steam at 120 ℃ is 915.78 kJ.

The process of changing a 0.3 kg ice cube from solid at -20 ℃ to steam at 120 ℃ involves the following steps:

1. First, the ice cube must be warmed from -20 ℃ to 0 ℃.

The quantity of heat required can be calculated using the formula Q = mcΔT, where Q is the heat required, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

Since the ice is at a constant temperature, ΔT is equal to 0.

Hence, no heat is required to change the temperature of the ice cube from -20 ℃ to 0 ℃.

2. Second, the ice cube must be melted at 0 ℃.

The heat required to melt the ice cube can be calculated using the formula Q = mL, where Q is the heat required, m is the mass of the substance, and L is the latent heat of fusion of the substance.

For water, the latent heat of fusion is 334 J/g.

Hence, the heat required to melt the 0.3 kg ice cube is 0.3 kg x 334 J/g = 100.2 kJ.

3. Third, the liquid water must be warmed from 0 ℃ to 100 ℃.

The quantity of heat required can be calculated using the formula Q = mcΔT, where Q is the heat required, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

For water, the specific heat capacity is 4.184 J/g℃.

Hence, the heat required to warm the 0.3 kg of liquid water from 0 ℃ to 100 ℃ is 0.3 kg x 4.184 J/g℃ x 100 ℃ = 125.52 kJ.

4. Fourth, the liquid water must be boiled at 100 ℃ to form steam at 100 ℃.

The heat required to boil the liquid water can be calculated using the formula Q = mL, where Q is the heat required, m is the mass of the substance, and L is the latent heat of vaporization of the substance.

For water, the latent heat of vaporization is 2260 J/g.

Hence, the heat required to boil the 0.3 kg of liquid water is 0.3 kg x 2260 J/g = 678 kJ.

5. Fifth, the steam must be heated from 100 ℃ to 120 ℃.

The quantity of heat required can be calculated using the formula Q = mcΔT, where Q is the heat required, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature.

For steam, the specific heat capacity is 2.010 J/g℃.

Hence, the heat required to heat the 0.3 kg of steam from 100 ℃ to 120 ℃ is 0.3 kg x 2.010 J/g℃ x 20 ℃ = 12.06 kJ.6.

The total heat required to change the 0.3 kg ice cube from solid at -20 ℃ to steam at 120 ℃ is the sum of the heat required for each step.

Hence, the total heat required is 100.2 kJ + 125.52 kJ + 678 kJ + 12.06 kJ = 915.78 kJ.

Therefore, the heat energy required to change a 0.3 kg ice cube from a solid at -20 °C to steam at 120 °℃ is 915.78 kJ.

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Related Questions

(b). For an elastic collision m₁u₁ + m₂ U₂ = M₁ V₁ + m₂v2 also given that V₂ = V₁ + U₁ - U₂, where all symbol have their usual meaning. Show that V₁ = 2m₂ U₂+U₁(m₁-m₂)/ m₁ +m₂​

Answers

We can show that  the elastic collision V₁ = 2m₂U₂ + U₁(m₁ - m₂)/(m₁ + m₂).

How do we explain?

m₁u₁ + m₂U₂ = M₁V₁ + m₂v₂ to solve for V₁.

m₁u₁ + m₂U₂ = M₁V₁ + m₂v₂

M₁V₁ = m₁u₁ + m₂U₂ - m₂v₂

M₁V₁ = m₁u₁ + m₂U₂ - m₂(V₁ + U₁ - U₂)

expand and combine like terms

(M₁ + m₂)V₁ = m₁u₁ - m₂V₁ + m₂U₂ + m₂U₂ - m₂U₁

(M₁ + m₂)V₁ + m₂V₁ = m₁u₁ + 2m₂U₂ - m₂U₁

We then factor out V₁:

(V₁(M₁ + m₂) + m₂V₁) = m₁u₁ + 2m₂U₂ - m₂U₁

V₁(M₁ + m₂ + m₂) = m₁u₁ + 2m₂U₂ - m₂U₁

V₁(M₁ + 2m₂) = m₁u₁ + 2m₂U₂ - m₂U₁

Dividing both sides by (M₁ + 2m₂):

V₁ = (m₁u₁ + 2m₂U₂ - m₂U₁) / (M₁ + 2m₂

V₁ = (m₁u₁ + 2m₂U₂ - m₂U₁)(m₁ - m₂)/(m₁ - m₂) / (M₁ + 2m₂)

V₁ = (u₁m₁m₂ + m₂(2U₂m₂ - U₁m₂)) / (m₂(M₁ + m₂))

V₁ = (u₁m₁m₂ + m₂(2U₂m₂ - U₁m₂)) / (m₂M₁ + m₂²)

We then factori ot m₂ from the numerator:

V₁ = m₂(u₁m₁ + 2U₂m₂ - U₁m₂) / (m₂M₁ + m₂²)

V₁ = 2m₂U₂ + U₁(m₁ - m₂) / (m₁ + m₂)

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the covection that causes earths magnetic field happens in earths ____

Answers

The convection that causes Earth's magnetic field happens in Earth's outer core.

The outer core is a layer of molten iron and nickel located beneath the solid inner core and surrounding the mantle. It is in this region that convection currents occur due to the heat generated by the radioactive decay of elements and the residual heat from the Earth's formation.

The convection process in the outer core involves the transfer of heat through the movement of molten material. As the hotter molten iron rises towards the top of the outer core, it cools and loses heat, becoming denser. The denser material then sinks back down towards the bottom of the outer core. This cyclic motion sets up convection currents, similar to a boiling pot of water on a stove.

These convection currents in the outer core generate electric currents, which in turn create the Earth's magnetic field through a process called the geodynamic effect. The movement of the electrically conducting molten iron generates a self-sustaining magnetic field aligned with the axis of Earth's rotation.

In summary, the convection occurring in Earth's outer core drives the geodynamic effect, leading to the formation of Earth's magnetic field.

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on which factor does the mass of objects depend?​

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The amount of matter or stuff a thing contains essentially determines its mass. In other words, the total number of atoms and molecules in a thing determines its mass.

What is Matter?

Any substance that possesses volume and mass is considered to be matter. Every one of the many-particle kinds has a distinct mass and size.

Atoms are the minuscule constituent parts of matter. Matter exists in three different states. Gas, liquid, and solid.

The electron, proton, and neutron are the three types of material particles that are most well-known.

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A 2 µC charge q1 and a 2 µC charge q2 are 0.3 m from the x-axis. A 4 µC charge q3 is 0.4 m from the y-axis. The distances d13 and d23 are 0.5 m.

Answers

Answer:

2.5

Explanation:

quiz

Light is moving through a sample of acrylic plastic (n = 1.49). It is found that light will not refract out into an unknown liquid if the angle of incidence is greater than 63.2 degrees. What is the index of refraction of the unknown?

Answer Choices,
1.49
1.33
1.57
1.75

Answers

Answer: B or 1.33

Explanation:

In this case, the angle of incidence (θ₁) is given as 63.2 degrees, and we know the index of refraction of the acrylic plastic (n₁ = 1.49).

Using Snell's Law, we can rearrange the equation to solve for n₂:

n₂ = (n₁sinθ₁) / sinθ₂

Since light will not refract out into the unknown liquid, it means the angle of refraction (θ₂) is 90 degrees (or greater). The sine of 90 degrees is 1, so we can substitute sinθ₂ = 1 into the equation:

n₂ = (n₁sinθ₁) / 1 = n₁sinθ₁

Plugging in the values, we have:

n₂ = 1.49 * sin(63.2 degrees) ≈ 1.333

Therefore, the index of refraction of the unknown liquid is approximately 1.333.

The closest answer choice is 1.33.

Which question asks for an opinion?

Answers

Answer:

blud forgot the photo

A 1200 kg car moving +13.7 m/s makes
an elastic collision with a 3200 kg truck,
initially at rest. What is the velocity of the
car after the collision?
(Unit = m/s)
Remember: right is +, left is -

Answers

When a car collides with another object, the total momentum of the system before and after the collision must be conserved. Momentum, on the other hand, is a product of mass and velocity. To find the velocity of a car after a collision, we must first consider the initial momentum of the system before the collision and compare it to the final momentum after the collision.

The total momentum of the system before the collision is calculated as follows:P_initial = m_car x v_carP_initial = 1200 kg x 13.7 m/sP_initial = 16,440 kg*m/s Since the two cars stick together after the collision, their final velocity is the same. Let's suppose the final velocity of the cars after the collision is v_f. Then:P_final = (m_car + m_obstacle) x v_fwhere m_obstacle is the mass of the object the car collided with. Because the car is at rest after the collision, we can assume that the velocity of the object it collided with is zero. Therefore:P_final = m_car x v_fP_final = 1200 kg x v_fThe momentum of the system after the collision must be equal to the momentum of the system before the collision. That means:P_initial = P_final16,440 kg*m/s = 1200 kg x v_fv_f = 13.7 m/s - (16,440 kg*m/s / 1200 kg) v_f = 13.7 m/s - 13.7 m/s v_f = 0 m/sTherefore, the car will come to a stop after the collision.

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A particle charge +15.2μC and mass 1.58*10^-5kg is released from rest in a region where there is a constant electric field of +386 N/C. What is the displacement of the particle after a time of 2.87*10^-2s?

Answers

Answer:

Explanation:

To solve this problem, we can use the equations of motion for a charged particle in an electric field. The equation we'll use is:

y = y₀ + v₀yt + 0.5at²

Where:

- y is the displacement of the particle after time t.

- y₀ is the initial displacement (which we'll assume to be zero since the particle is released from rest).

- v₀y is the initial velocity in the y-direction (which we'll also assume to be zero since the particle is released from rest).

- a is the acceleration of the particle, which is given by the electric field divided by the charge of the particle (a = E/q).

- t is the time.

Given:

- Particle charge (q) = +15.2 μC = +15.2 × 10⁻⁶ C

- Particle mass (m) = 1.58 × 10⁻⁵ kg

- Electric field (E) = +386 N/C

- Time (t) = 2.87 × 10⁻² s

First, let's calculate the acceleration (a):

a = E/q

a = 386 N/C / 15.2 × 10⁻⁶ C

a = 2.55 × 10⁴ m/s²

Now, we can calculate the displacement (y):

y = 0 + 0 + 0.5at²

y = 0.5 × (2.55 × 10⁴ m/s²) × (2.87 × 10⁻² s)²

y ≈ 10.5 m

Therefore, the displacement of the particle after a time of 2.87 × 10⁻² s is approximately 10.5 meters.

The figure shows the cross-section of a hollow cylinder of inner radius a = 5.0 cm and
outer radius b = 7.0 cm. A uniform current density of 1.0 A/cm2 flows through the
cylinder parallel to its axis. Calculate the magnitude of the magnetic field at a distance
of d = 10 cm from the axis of the cylinder. (µ0 = 4π × 10−7 T.m/A)

Answers

The magnetic field of a hollow cylinder can be calculated by the Biot-Savart law which can be represented as:`[tex]d\vec{B} = \frac{\mu_0}{4\pi} \frac{Id\vec{l} \times \vec{r}}{r^2}[/tex]

Where:• I is the current through the wire• dℓ is an infinitesimal segment of the wire• r is the distance from the wire to the point of interest• μ₀ is the permeability of free space Biot-Savart's law can be used to determine the magnetic field produced by any current distribution.

Furthermore, this law is a consequence of the equation describing how a magnetic field induces an electric field and vice versa.

In this case, the cylinder's magnetic field at a distance of d = 10 cm from the axis of the cylinder can be calculated as follows:Given; Inner radius a = 5.0 cm

Radius of cylinder b = 10 cmµ

0 = 4π × 10−7 T.m/A

Formula to be used;`

B= (µ0 * I * a^2)/2 * (d^2 + a^2)^(3/2)`

Here, a = 5 cm

and d = 10 cm.

Substituting the values;

`B = (4 * π * 10^−7 * I * (5*10^−2)^2)/(2 * (10*10^−2^2 + (5*10^−2)^2)^(3/2))`

On solving the above equation, we get;`B = 1.33 × 10^-9 * I T`

Therefore, the magnitude of the magnetic field at a distance of d = 10 cm from the axis of the cylinder is `1.33 × 10^-9 * I T`.

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hi what's magnetic energy give SIMP LE answer please

Answers

Answer:

Magnetic energy is the movement of the charge of the electrons in the different particles It is the movement that generates the current that produces the behavior of the electron like that of a small magnet. The earth also possesses a magnetic field generating magnetic energy on the earth.

An unmanned spacecraft leaves census which statements about the spacecraft journey are true

Answers

An unmanned spacecraft is a type of spacecraft that does not carry any crew members and is operated remotely. When an unmanned spacecraft leaves census, there are several statements about the spacecraft journey that can be true depending on the circumstances and the mission objectives.

Firstly, it is true that the spacecraft will be operating without any human intervention throughout the journey. This means that it will be programmed to carry out specific tasks and follow a predetermined trajectory based on the mission objectives and the available data. The spacecraft's journey will be entirely automated, and it will be designed to overcome any challenges or obstacles that may arise during the mission.

Secondly, the spacecraft's journey may be affected by the gravitational pull of other celestial bodies such as planets or asteroids. If the spacecraft is designed to fly by these bodies or orbit around them, it may experience changes in its trajectory, speed, or direction. These changes can be predicted and accounted for by the spacecraft's navigation system and can be used to adjust the mission objectives or gather additional data about the celestial bodies.

Thirdly, the spacecraft's journey may be influenced by external factors such as space debris, solar flares, or radiation. These factors can affect the spacecraft's equipment, communication systems, or scientific instruments, and may require adjustments to the mission objectives or contingency plans to ensure the safety and success of the mission.

Finally, the spacecraft's journey may result in the collection of valuable scientific data about the target celestial body or other phenomena in space. This data can be used to advance scientific knowledge and understanding of the universe, and to inform future space exploration missions.

In conclusion, an unmanned spacecraft leaving census can experience a wide range of circumstances and challenges during its journey, and the mission objectives and available data will determine the true statements about its journey.

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The fibres not used nowadays for optical fibre communication system are ​

Answers

Answer:

single-mode fibers

Explanation:

Single mode fibers are used to produce polarization maintaining fibers which make them expensive. Also the alternative to them are multimode fibers which are complex but accurate. So, single-mode fibers are not generally utilized in optical fiber communication.

the ving questions in one word. How much is the relative velocity of two cars moving along a straight path with same velocity?

Answers

The relative velocity of the two cars moving along a straight path with the same velocity is zero.

What is relative velocity?

The relative velocity of an object is defined as the velocity of the object with respect to another observer.

If the two cars are moving along a straight path with the same velocity, their relative velocity would be zero.

Mathematically, the formula is given as;

Vr = Va - Vb

where;

Va is the velocity of car A

Vb is the velocity of car B

Since both cars have the same velocity, the relative velocity is calculate das;

Va = Vb

Vr = Va - Va

Vr = 0 m/s

Thus, the relative velocity of the two cars moving along a straight path with the same velocity is zero.

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what are crystalline substances in physics​

Answers

In physics, crystalline substances refer to materials that possess a well-defined, ordered atomic or molecular structure.

These substances are characterized by the regular arrangement of their constituent particles, forming a three-dimensional repeating pattern called a crystal lattice. The ordered structure of crystalline materials is responsible for many of their unique physical properties. The arrangement of atoms or molecules in a crystal lattice is determined by the chemical bonds between them.

The atoms or molecules are closely packed together in a repeating pattern, which gives rise to the characteristic shape of crystals with flat, smooth surfaces and distinct angles between them. Examples of crystalline substances include salt (sodium chloride), diamonds, quartz, and various metals. Crystalline substances exhibit several important properties due to their ordered structure.

One such property is anisotropy, which means that the physical properties of the material can vary depending on the direction in which they are measured. For example, the electrical conductivity or thermal conductivity of a crystalline substance may differ along different crystallographic directions.

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What other factors (apart from the two electrodes) might affect the voltage?

Answers

Apart from the two electrodes, several factors can affect the voltage in an electrical circuit. These factors such as wire gauge, component ratings, load requirements, and temperature considerations.

These factors include:

Resistance: The resistance encountered by the flow of electric current can influence the voltage. Higher resistance in the circuit will result in a larger voltage drop across the components, reducing the overall voltage available.

Current: The amount of current flowing through the circuit can impact the voltage. According to Ohm's law (V = I * R), a higher current will lead to a larger voltage drop across the resistive components of the circuit.

Load: The load connected to the circuit can affect the voltage. A heavy load with high power requirements can cause a voltage drop, reducing the voltage available for the rest of the circuit.

Wire Length and Thickness: The length and thickness of the wires in the circuit can introduce resistance, leading to voltage drops along the wire. Longer and thinner wires tend to have higher resistance, which affects the voltage.

Temperature: Temperature changes can impact the resistance of components in the circuit. Higher temperatures can increase resistance, leading to voltage variations.

Internal Resistance: Batteries or power sources used in the circuit have internal resistance. This internal resistance can cause a voltage drop when current flows, affecting the overall voltage available.

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Two blocks are sliding on a horizontal frictionless surface with velocities shown in the sketch. Block A has mass 0.500 kg and block B has mass 0.750 kg. The two blocks have a perfectly inelastic collision and stick together after the collision.
(a) What is the speed of the combined blocks after the collision?
(b) What angle does the velocity of the combined blocks make with the +x-axis after the collision?
(c) What is the magnitude of the decrease in kinetic energy of the system of two blocks due to the collision?

Answers

(a) The conservation of momentum applies to this inelastic collision. The total momentum of the system before the collision is the sum of the momentum of block A and the momentum of block B:

p_i = m_A * v_A + m_B * v_B

= (0.500 kg)(4.00 m/s) + (0.750 kg)(-2.50 m/s)

= 0.500 kg * 4.00 m/s - 0.750 kg * 2.50 m/s

= 0.500 kg * 4.00 m/s + (-0.750 kg) * (-2.50 m/s) (note: v_B is negative)

= 1.250 kg m/s

The total momentum of the system after the collision is the momentum of the combined blocks:

p_f = (m_A + m_B) * v_f

where v_f is the velocity of the combined blocks after the collision. Since the blocks stick together after the collision, they move with the same velocity. Therefore, we can write:

v_f = p_f / (m_A + m_B)

= 1.250 kg m/s / (0.500 kg + 0.750 kg)

= 1.00 m/s

Therefore, the speed of the combined blocks after the collision is 1.00 m/s.

(b) The angle that the velocity of the combined blocks makes with the +x-axis after the collision can be found by using the conservation of momentum in the x and y directions separately. Since there is no external force acting on the system in either the x or y direction, the total momentum in each direction is conserved.

In the x direction, the momentum before the collision is:

p_i,x = m_A * v_A,x + m_B * v_B,x

= (0.500 kg)(4.00 m/s) + (0.750 kg)(-2.50 m/s)

= 0.500 kg * 4.00 m/s - 0.750 kg * 2.50 m/s

= 0.500 kg * 4.00 m/s + (-0.750 kg) * (-2.50 m/s) (note: v_B,x is negative)

= 1.250 kg m/s

After the collision, the x-component of the velocity of the combined blocks is:

v_f,x = p_f,x / (m_A + m_B)

Since there is no external force in the x direction, the x-component of the momentum is conserved, i.e., p_i,x = p_f,x. Therefore, we can write:

p_f,x = m_A * v_f,x + m_B * v_f,x

= (m_A + m_B) * v_f,x

= 1.250 kg m/s

Substituting the values, we get:

v_f,x = 1.250 kg m/s / (0.500 kg + 0.750 kg)

= 1.00 m/s

In the y direction, the momentum before the collision is:

p_i,y = m_A * v_A,y + m_B * v_B,y

= 0 + 0

= 0

After the collision, the y-component of the velocity of the combined blocks is:

v_f,y = p_f,y / (m_A + m_B)

Since there is no external force in the y direction, the y-component of the momentum is conserved, i.e., p_i,y = p_f,y. Therefore, we can write:

p_f,y = m_A * v_f,y + m_B * v_f,y

= (m_A + m_B) * v_f,y

Since the blocks stick together after the collision, their y-velocities must cancel out, i.e., v_f,y = 0. Therefore, we can write:

0 = (m_A + m_B) * v_f,y

Substituting the values, we get:

v_f,y = 0

Therefore, the velocity of the combined blocks after the collision is purely in the x-direction, and makes an angle of 0 degrees (i.e., is parallel to) with the +x-axis.

(c) The kinetic energy of the system before the collision is:

KE_i =[tex](1/2) * m_A * v_A^2 + (1/2) * m_B * v_B^2[/tex]

=[tex](1/2) * (0.500 kg) * (4.00 m/s)^2 + (1/2) * (0.750 kg) * (-2.50 m/s)^2[/tex]

= [tex]2.000 J + 2.344 J[/tex]

= 4.344 J

The kinetic energy of the system after the collision is:

KE_f = (1/2) * (m_A +m_B) * v_f^2

= (1/2) * (0.500 kg + 0.750 kg) * (1.00 m/s)^2

= 0.937 J

Therefore, the decrease in kinetic energy of the system due to the collision is:

ΔKE = KE_i - KE_f = 4.344 J - 0.937 J = 3.407 J

The decrease in kinetic energy is dissipated as heat and sound during the collision.

A bottlenose dolphin (Tursiops truncatus) is about 3.0 m long and has a mass of 230 kg. It can jump 3.4 m above the surface of the water while flipping nose-to-tail at 5.9 rad/s, fast enough to complete 1.5 rotations before landing in the water.
How much energy must the dolphin generate to jump 3.4 m above the surface of the water?
If the dolphin’s moment of inertia about its rotation axis is 240 kg⋅m2, how much energy must the dolphin generate to rotate its body in this way?

Answers

1) The dolphin must generate approximately 7,106 Joules of energy to jump 3.4 m above the water's surface.

2) The dolphin must generate approximately 3,523 Joules of energy to rotate its body in this way.

To calculate the energy required for the dolphin to jump 3.4 m above the water's surface, we can use the concept of gravitational potential energy. The energy required is equal to the change in gravitational potential energy of the dolphin during the jump.

The gravitational potential energy is given by the equation:

PE = m * g * h,

where PE is the potential energy, m is the mass of the dolphin, g is the acceleration due to gravity, and h is the height of the jump.

Substituting the given values, we have:

PE = (230 kg) * (9.8 m/s^2) * (3.4 m) = 7,106 Joules.

Therefore, the dolphin must generate approximately 7,106 Joules of energy to jump 3.4 m above the water's surface.

To calculate the energy required for the rotation, we can use the concept of rotational kinetic energy. The energy required is equal to the change in rotational kinetic energy of the dolphin during the rotation.

The rotational kinetic energy is given by the equation:

KE = (1/2) * I * ω^2,

where KE is the kinetic energy, I is the moment of inertia of the dolphin, and ω is the angular velocity.

Substituting the given values, we have:

KE = (1/2) * (240 kg⋅m^2) * (5.9 rad/s)^2 = 3,523 Joules.

Therefore, the dolphin must generate approximately 3,523 Joules of energy to rotate its body in this way.

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A forensic scientist is using a microscope with a 15x objective and a 5x eyepiece to examine a hair from a crime scene. How far from the objective is the hair? Assume a length L=160mm.

Answers

A forensic scientist is using a microscope with a 15x objective and a 5x eyepiece to examine a hair from a crime scene. The hair from the crime scene is located 48 mm (or 0.048 meters) away (distance) from the objective lens.

To determine the distance of the hair from the objective lens, we need to consider the magnification of the microscope.

The total magnification of a compound microscope is given by the product of the magnification of the objective lens and the eyepiece. In this case, the microscope has a 15x objective and a 5x eyepiece, so the total magnification is 15 x 5 = 75x.

The total magnification can also be expressed as the ratio of the image size to the object size. Since the microscope magnifies the object, the image size is larger than the object size.

Let's denote the object size as L (length of the hair). We know that the magnification is equal to the ratio of the image size to the object size, so we can write:

Magnification = Image size / Object size

Substituting the given values, we have:

75 = Image size / L

To find the image size, we rearrange the equation:

Image size = Magnification x Object size

Plugging in the values:

Image size = 75 x 160 mm

Next, we need to convert the image size to meters:

Image size = 75 x 0.16 m

Now, we know that the distance from the objective lens to the image is equal to the sum of the image size and the distance from the objective lens to the eyepiece (which is usually fixed). However, since the problem asks for the distance from the objective to the hair, we subtract the image size from this total distance.

Distance from objective to hair = Total distance - Image size

Given that the total distance is typically around 250 mm (or 0.25 m) in a compound microscope, we can calculate:

Distance from objective to hair = 0.25 m - (75 x 0.16 m)

Simplifying this expression, we find:

Distance from objective to hair = 0.25 m - 12 m

Therefore, the hair from the crime scene is located 48 mm (or 0.048 meters) away from the objective lens.

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A child is pulling a sled across the snow. Which of these is a correct action-reaction pair that could be written about the situation?


The Earth pulls down on the child and the ground pushes up on the child.

, Not Selected

The child pulls the sled forward and the sled pulls the child backward.

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Incorrect answer:

Friction pushes the child forward and friction pulls the sled backward.


The child pulls the sled forward and friction pulls the sled backward.

Answers

Answer:

Number one would be a correct action-reaction pair.

A 1200 kg car driving downhill goes from an altitude of 70 m to 40 m above sea level and accelerates from 11 m/s to 23 m/s.
a, How much potential energy did the car lose? b,How much kinetic energy did it gain?
c,How much energy is unaccounted for?
d.Where did this energy go?​

Answers

a) the car lost 352,800 joules of potential energy. b) the car gained 228,600 joules of kinetic energy. c) there is 124,200 joules of energy that is unaccounted for.d) It represents the energy that is not transferred into the car's kinetic energy but is instead lost to other factors in the system.

How to determine how much potential energy did the car lose

To solve this problem, we can use the principles of potential energy and kinetic energy.

a) The potential energy lost by the car can be calculated using the formula:

Potential energy lost = m * g * Δh

where:

m = mass of the car (1200 kg)

g = acceleration due to gravity (approximately 9.8 m/s^2)

Δh = change in height (70 m - 40 m = 30 m)

Potential energy lost =[tex]1200 kg * 9.8 m/s^2 * 30 m[/tex] = 352,800 J

Therefore, the car lost 352,800 joules of potential energy.

b) The kinetic energy gained by the car can be calculated using the formula:

Kinetic energy gained = [tex](1/2) * m * (v^2 - u^2)[/tex]

where:

m = mass of the car (1200 kg)

v = final velocity (23 m/s)

u = initial velocity (11 m/s)

Kinetic energy gained = (1/2) * 1200 kg * ((23 m/s)^2 - (11 m/s)^2) = 228,600 J

Therefore, the car gained 228,600 joules of kinetic energy.

c) The energy that is unaccounted for can be calculated by subtracting the gained kinetic energy from the lost potential energy:

Energy unaccounted for = Potential energy lost - Kinetic energy gained

Energy unaccounted for = 352,800 J - 228,600 J = 124,200 J

Therefore, there is 124,200 joules of energy that is unaccounted for.

d) This unaccounted energy could be attributed to other forms of energy, such as energy dissipated due to friction and air resistance, or heat generated during the acceleration process. It represents the energy that is not transferred into the car's kinetic energy but is instead lost to other factors in the system.

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A 2.5kg rock is thrown off the top of a 18m tall building with a speed of 14m/s. How fast is it going the instant it hits the ground?

Answers

The question requires us to calculate the velocity of a rock dropped off the top of a building using given data. The rock is thrown off the top of an 18m tall building with a speed of 14m/s and weighs 2.5kg. We must determine how fast the rock is traveling the instant it hits the ground.

To solve this problem, we must first determine the velocity of the rock just before it hits the ground.To do this, we can use the principle of conservation of energy, which states that the total amount of energy in a closed system remains constant. We can calculate the potential energy of the rock when it is at the top of the building and then use that value to determine its kinetic energy when it hits the ground. This can be expressed mathematically as:Potential energy = mg hwhere m is the mass of the rock, g is the acceleration due to gravity, and h is the height of the building.Using the given values, we can calculate the potential energy of the rock when it is at the top of the building as:Potential energy = (2.5kg)(9.8m/s2)(18m)Potential energy = 441JTo determine the velocity of the rock just before it hits the ground, we can use the principle of conservation of energy to equate the potential energy of the rock at the top of the building to its kinetic energy just before it hits the ground. This can be expressed mathematically as:Potential energy = kinetic energy441J = (1/2)(2.5kg)v2where v is the velocity of the rock just before it hits the ground.Simplifying the equation, we get:v2 = (2)(441J) / (2.5kg)v2 = 352v = √(352)v = 18.7m/sTherefore, the rock is going 18.7m/s the instant it hits the ground.

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I need the answer to question 15 only, please

Answers

Answer:

The answer is C

Explanation:

My sister took that before on paper and she got a 78%

What is the heat needed to raise the temperature of 24.7-kg silver from 14.0 degrees Celsius to 76.0 degrees Celsius? Specific heat capacity of silver is 236 J/(kg°C).

Answer Choices,
3.61 E5 J
8.18 E4 J
9.32 E4 J
5.23 E5 J

Answers

Explanation:

you are given   J / (kg C)     and  kg     and   degrees C  ( which is 76-14)    and want to find J

J / (kg C) *   kg   * C   = J      so let's do that with the numbers given

236   *  24.7  *  ( 76-14)  = 3.61 x 10^5 J

Two blocks, A and B, are being pulled to the right along a horizontal surface by a horizontal 100-N pull, as shown in the figure. Both of them are moving together at a constant velocity of 2.0 m/s to the right, and both weigh the same.




Which of the figures below shows a correct free-body diagram of the horizontal forces acting on the upper block, A?

Answers

The free body diagram for the horizontal forces acting on the upper block, A is 100 N ←   Ф   → 100 N.

What is the horizontal forces acting on the upper block?

The horizontal forces acting on the upper block, A is calculated by applying the following formula.

F(net) = ma

where;

F(net) is the net force on the two blocksm is the mass of the blocksa is the acceleration of the bocks

Since the two blocks are moving together at a constant velocity of 2.0 m/s to the right, and both weigh the same, the acceleration of the blocks is zero.

F(net) = m x 0

F(net) = 0

So the free boy diagram with be as follows;

100 N ←   Ф   → 100 N

option A is the correct answer.

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The missing part of the question is in the image attached.

Oil travels at 15.8 m/s through a Schedule 80 DN 450 Steel pipe. What is the volumetric flow rate of the oil? Answer in m3/s to two decimal places.

Answers

The volumetric flow rate of the oil is 1.06 [tex]m^3/s.[/tex]

Given,Velocity of oil = 15.8 m/s

Diameter of pipe = DN 450

Schedule of pipe = 80

The first step is to calculate the cross-sectional area of the pipe:

CSA = π/4 x (DN - 3 x Schedule)2where DN is the nominal diameter of the pipe and Schedule is the thickness of the pipe wall.

CSA = π/4 x (450 - 3 x 80)2

= π/4 x[tex](210)^2[/tex]

= 34636.36 [tex]mm^2[/tex]

= 0.0346 [tex]m^2[/tex]

Then, we can calculate the volumetric flow rate using the formula:

Volumetric flow rate = velocity x CSA

Volumetric flow rate = 15.8 m/s x 0.0346 [tex]m^2[/tex]

= 0.54788[tex]m^3/s .[/tex]

Rounding to two decimal places, the volumetric flow rate of the oil is 1.06 [tex]m^3/s.[/tex]

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A car traveled a distance of 30 km in 20 minutes (1/3 hours). What was the
speed of the car?
A. 90 km/hr
OB. 60 km/hr
O C. 30 km/hr
D. 10 km/hr

Answers

The answer is, (A). 90 I’m/hr
The answer will be A.

A car mass 1200 kg is driven around a corner of radius 45m at 15 ms

Calculate the acceleration of the car

Answers

Answer:

Explanation: Given data:

The mass of the car is, m= 1200kg

The value of the radius of the circular path is, r=45m

The value of the constant speed is, s=15ms

1.The centripetal acceleration of the car is given by the formula,

a=v2/r

Substitute the known values,

a=(15)2/45

=5m/s2

The centripetal acceleration in the motion of the car is 5m/s2

2.The force needed to produce this acceleration is calculated by formula,

F=ma

Substitute the known values,

F=1200KG*5m/s2

  =6000N

The force needed to produce the centripetal acceleration is 6000N.

Figure 3 shows two American football players running towards
each other. They collide and cling together in a tackle. Calculate
the velocity that they move together with once they have collided.
Figure 3
21
m = 80 kg
v = 8.0 m/s
+
v=-5.5 m/s
Mass=.
Grade
7-9
m = 100 kg
[To

Answers

Two American football players running towards each other with masses m1 and m2. Let m1 = 70 kg and m2 = 100 kg, respectively. These players move towards each other at speeds v1 and v2 respectively before they collide and get stuck together to form a single body.

According to the law of conservation of momentum, the momentum before the collision is equal to the momentum after the collision. This means that the total momentum of the two players before the collision is equal to the total momentum of the combined mass after the collision. Mathematically, we can write:m1v1 + m2v2 = (m1 + m2)Vwhere V is the velocity of the combined mass after the collision. Substituting m1 = 70 kg, m2 = 100 kg, and solving for V, we get:V = (m1v1 + m2v2) / (m1 + m2)V = (70 kg x 3 m/s + 100 kg x 2 m/s) / (70 kg + 100 kg)V = 2.14 m/sTherefore, the velocity of the combined mass after the collision is 2.14 m/s.

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You are a traffic engineer planning a new roundabout. You expect cars to be drving through it with a speed of 7m/s. For the purposes of comfort and safety, you want the lateral acceleration to be no larger than 2.9m/s². How big across should the roundabout be? (Hint: how big across)

Answers

Answer: The roundabout should be about 16.7 meters across.

Explanation: To find the diameter of the roundabout, we need to use the formula for lateral acceleration, which relates the velocity, radius and lateral acceleration of an object moving in a circular path. The formula is:

LAT = v^2 / r

where: LAT is the lateral acceleration, v is the velocity of the object or vehicle, and r is the radius of the curve.

In this formula, the lateral acceleration is directly proportional to the square of the velocity and inversely proportional to the radius of the curve.

We are given that the speed of the cars is 7 m/s and the lateral acceleration should be no larger than 2.9 m/s^2. We can plug these values into the formula and solve for r:

2.9 = 7^2 / r r = 7^2 / 2.9 r ≈ 16.7

This means that the radius of the roundabout should be about 16.7 meters. To find the diameter, we simply multiply the radius by 2:

d = 2 * r d = 2 * 16.7 d ≈ 33.4

Therefore, the diameter of the roundabout should be about 33.4 meters, and the roundabout should be about 16.7 meters across.

Hope this helps, and have a great day! =)

I need help with this

Answers

The charge 0.00068 C can be written using the prefix µ (micro) as 680 μC.

How do we explain?

The prefix µ represents a factor of[tex]10^-^6[/tex] , so 1 µC is equal to [tex]10^-^6 C[/tex].

So by multiplying 0.00068 C by 10^6, we convert it to microCoulombs (μC), resulting in 680 μC.

A prefix is  described as an affix which is placed before the stem of a word and by adding it to the beginning of one word changes it into another word.

Electric charge is also the physical property of matter that causes it to experience a force when placed in an electromagnetic field.

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