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The actual weight of the gas = apparent weight + weight.

The actual weight = [tex]W_{A}[/tex] + W

Given that a plastic bag is massed. It is then filled with a gas which is insoluble in water and massed again.

If the apparent weight of the gas is the difference between these two masses, then let the apparent weight = [tex]W_{A}[/tex]

The gas is squeezed out of the bag to determine its volume by the displacement of water. Since

density = mass / volume

The density of water is 1000 kg/[tex]m^{2}[/tex]

we can get the mass of the gas by making m the subject of the formula.

W = mg

The actual weight of the gas = apparent weight + weight

That is,

The actual weight =  [tex]W_{A}[/tex] + W

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The time taken by  the projectile to reach its maximum height is 0.62 second.

Given parameters:

Horizontal component of initial velocity of the projectile: [tex]u_x[/tex] = 6.1 m/s.

To reach maximum height, the projectile should be throwed in 45 degree. So, vertical component of initial velocity of the projectile: [tex]u_y[/tex] = 6.1 m/s.

Given,  The acceleration of gravity is 9.8 m/s².

At maximum height the vertical component of velocity of the projectile becomes zero due to acceleration due to gravity acts downwards.

So, from formula [tex]v_y[/tex] = [tex]u_y[/tex] - gt; we can write:

0 = 6.1 - 9.8t

⇒ t = 6.1/9.8 = 0.62 s.

Hence, the time taken by  the projectile to reach its maximum height is 0.62 second.

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Acceleration = (change in velocity ( final speed - starting speed))/ (time)

Acceleration = (18-30)/10.5

Acceleration = -12/10.5

Acceleration = -1.14 m/s^2

Distance = 30m/s x 10.5s + 1/2(1.14)(10.5)^2

Distance = 252.2 meters

Answer: An ice hockey player picks up a trophy as he slides past it.

Two pool balls colliding precisely inelastically and rebounding off one another is an example. So, the correct option is A.

Inelastic collisions are those in which the total kinetic energy is lower after the impact than it was before. The stick is travelling quickly in the direction of the ball before to the contact. The stick comes to a stop following the accident. It transfers some of its kinetic energy to the cue ball, which rolls forward.

The type of collision mentioned in the given example is known as an inelastic collision is one in which the system's momentum is preserved but its kinetic energy is not. In a ballistic pendulum, an example of an inelastic collision. Dropped ball of clay is another illustration of an inelastic collision.

Thus, the correct option is A.

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Your question is incomplete, most probably the complete question is:

Which of the following statements describes a perfectly inelastic collision?

A. Two billiard balls bounce off of each other,

B. A car crashes into a tree and rebounds in the other direction,

C. A wad of chewing gum is thrown and sticks to a moving truck,

D. Two ice skaters hit each other and fall over in opposite directions,

Answer:

Ammonium perchlorate NH4ClO4

Ammonium Nitrate

Calcium Cyanamide

When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.

Hopefully this helps :)

Answer:

less

Explanation:

heat travels from hotter to colder object. hotter object has more kinetic energy.

Let m₁ and m₂ be the masses of the two objects, and v₁ and v₂ their initial velocities. So

m₁ = 30.0 g = 0.0300 kg

m₂ = 13.0 g = 0.0130 kg

v₁ = + 20.5 cm/s = 0.205 m/s

v₂ = + 15.0 cm/s = 0.150 m/s

and we want to find v₁' and v₂', the final velocities of either object after their collision.

Momentum is conserved throughout the objects' collision, so that

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

where v₁' and v₂' are the first and second object's velocities after the collision.

Kinetic energy is also conserved, so that

1/2 m₁v₁² + 1/2 m₂v₂² = 1/2 m₁(v₁')² + 1/2 m₂(v₂')²

or

m₁v₁² + m₂v₂² = m₁(v₁')² + m₂(v₂')²

From the first equation (omitting units), we have

0.0300 • 0.205 + 0.0130 • 0.150 = 0.0300 v₁' + 0.0130 v₂'

0.0810 = 0.0300 v₁' + 0.0130 v₂'

81 = 30 v₁' + 13 v₂'

From the second equation,

0.0300 • 0.205² + 0.0130 • 0.150² = 0.0300 (v₁')² + 0.0130 (v₂')²

0.00155 ≈ 0.0300 (v₁')² + 0.0130 (v₂')²

1.55 ≈ 30 (v₁')² + 13 (v₂')²

Solving both equations simultaneously gives two solutions, one of which corresponds to the initial conditions. The other yields

v₁' ≈ + 0.172 m/s

and

v₂' ≈ + 0.227 m/s

The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.

The given parameters;

The normal force on the wooden box is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]

The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;

[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]

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From the information provided in the question, the mass of the beaker is 144.4 g.

From the information provided in the complete question;

volume of water = 74 mL

Mass of water = 74 g

specific heat of glass = 0.18 cal/g ∙ °C

specific heat of water = 1.0 cal/g ∙ C°

Mass of glass =  x g

Total heat gained by the system = 2000.0cal

Temperature rise = 20.0°C

Heat gained by system = Heat gained  by glass + Heat gained by water

Heat gained by glass = x ×  0.18 × 20

Heat gained by water = 74  ×  1.0 × 20

Hence;

2000 =  (x ×  0.18 × 20) + ( 74  ×  1.0 × 20)

2000 - 1480 =  (x ×  0.18 × 20)

x = 520/3.6

x = 144.4 g

Missing parts;

A glass beaker of unknown mass contains 74.0 ml of water. The system absorbs 2000.0cal of heat and the temperature rises 20.0°C as a result. What is the mass of the beaker? The specific heat of glass is 0.18

cal/g °C, and that of water is 1.0 cal/g °C.​

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Answer:

hi

Explanation:

i like saying hi

Rest in physics generally refers to the state in which an object is stationary.

Hence, in Physics, an object is in a state of rest when it does not move from one point to another.

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Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.

By Newton's second law, the net vertical force is

• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0

where a is the acceleration of the wagon.

Solve for n (the magnitude of the normal force) :

n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N

Then

f = 0.500 (58.0 N) = 29.0 N

Meanwhile, the horizontal component of the applied force has magnitude

(80.0 N) cos(30.0°) ≈ 69.3 N

Now calculate the work done by either force.

• friction: -(29.0 N) (10.0 m) = -290. J

• pull: (69.3 N) (10.0 m) = 693 J

Answer:

N1 sin theta1 = N2 sin theta2       Snell's Law

For the refracted ray to be reflected then sin theta2 = 1

N1 sin theta 1 = N2

Also N2 must be less than N1 for complete reflection

sin theta1 = N2/N1

If you are considering air and glass then N2 = 1   (for air)

sin theta1 = 1 / N2    where N2 must be for glass in this case

The acceleration due to gravity on the given planet is 8.71 m/s².

The given parameters;

The acceleration due to gravity on the planet is calculated by applying following formula as follows;

[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2 \pi } = \sqrt{\frac{l}{g}} \\\\\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\g = \frac{4\pi^2 l}{T^2} \\\\g = \frac{4 \times (\pi)^2 \times 0.45}{1.428^2} \\\\g = 8.71 \ m/s^2[/tex]

Thus, the acceleration due to gravity on the given planet is 8.71 m/s².

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Answer:

Our eyes are most sensitive to the wavelengths corresponding to the yellow and green colors of the spectrum. Flashy signs and some fire engines are painted in a yellowish-green color to attract our attention.

The distance between the glass to have the given heat loss is 2.54 m.

The given parameters:

The area of the glass window is calculated as follows;

[tex]A = 0.6 \times 0.9\\\\A = 0.54 \ m^2[/tex]

The distance between the glass is calculated as follows;

[tex]Q = \frac{KA \Delta T}{\Delta x} \\\\\Delta x = \frac{kA \Delta T}{Q} \\\\\Delta x = \frac{0.8 \times 0.54 \times 24 }{4.09} \\\\\Delta x = 2.54 \ m[/tex]

Thus, the distance between the glass to have the given heat loss is 2.54 m.

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This question involves the concepts of tension in a string and the wavelength of the wave in a string.

The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.

The speed of a wave on a string is given by the following formula:

[tex]v=\sqrt{\frac{T}{\mu}}[/tex]

where,

v = speed of wave = fλ

f = frequency of the wave

λ = wavelength of the wave

T = tension force

μ = linear mass density of the string

Therefore,

[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]

It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,

[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]

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Answer:

8.01e-22

Explanation:

The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

The extension produced by one mole of lead atom is calculated by applying Hooke's law;

F = kx

mg = kx

x = mg/k

x = (0.207 x 9.8) / (20)

x = 0.101 m

The Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;

E = ¹/₂kx²

E = ¹/₂ (20)(0.101)²

E = 0.102 J

Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

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Gravitational pull downward on the object:  7.8 x 9.8 = 76.44N

A. 7.8 x 1.5 = 11.7N upward force

Tension = 76.44 + 11.7 =  88.14 N

Answer: 88.14 N

B. 76.44 - 11.7 = 64.74N

Answer: 64.74 N

Answer:

88.14 N

64.74 N

Explanation:

The potential difference between the initial and final point is 3.0 V.

The given parameters:

The potential difference between the initial and final point is calculated as follows;

[tex]E = \frac{V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2}[/tex]

where:

[tex]d_2[/tex] is the distance of the electron between the positive and negative plate

[tex]0.1 + d_2 + 0.05 = 0.2\\\\d_2 + 0.15 = 0.2\\\\d_2 = 0.2 - 0.15\\\\d_2 = 0.05 \ m[/tex]

[tex]V_2 = \frac{V_1d_2}{d_1} \\\\V_2 = \frac{12 \times 0.05}{0.2} \\\\V_2 = 3.0 \ V[/tex]

Thus, the potential difference between the initial and final point is 3.0 V.

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A.

The work has a negative value.

Since the gas in the chamber expands, it increases in volume and does positive work on the piston since the change in volume is positive.

Work, W = pΔV. Since the gas expands, ΔV > 0. So, W > 0. Thus Work done by the gas is positive.

Since in ΔU = Q + W, W here is work done by the surroundings, W is negative since it is the opposite of the work done on the surroundings by the gas..

So, the work has a negative value.

B.

The heat added to the gas chamber is 17 J

From the first law of thermodynamics, ΔU = Q + W where ΔU = internal energy change = U₂ - U₁ where U₁ = initial energy = 8 J and U₂ = final energy = 30 J, Q = heat added to the gas chamber and W = work on piston = - 5 J

ΔU = Q + W

U₂ - U₁  = Q + W

30 J - 8 J = Q + (-5 J)

22 J = Q - 5 J

Q = 22 J + 5 J

Q = 27 J

The heat added to the gas chamber is 27 J.

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Hello!

We can use the following angular kinematic equation to solve:

α = Δω/Δt, or (ωf-ωi)/t

Plug in the given values:

α = (25 - 10)/5 = 15/5 = 3 rad/sec²

Answer:

The kinetic molecular theory of matter asserts that: Matter is made up of particles that are continually moving. All particles contain energy, however the energy fluctuates depending on the temperature the sample of matter is in. This in turn decides whether the material exists in the solid, liquid, or gaseous form.

Explanation:

Hope it helps:)

Answer:

option B is the correct answer

Explanation:

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5.08 m

Explanation:

The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:

[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]

where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get

[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]

Taking the square root of r^2, we then get the distance as

[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]

The values are given as follows:

[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]

[tex]g = 9.8\:\text{m/s}^2[/tex]

[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]

[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]

Putting in all of these values in our equation for r,

[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]

Hi there!

We know that:

Initial Total Mechanical Energy = Final Total Mechanical Energy

(Ei = Ef)

In this instance:

Ei = Gravitational Potential Energy

Ef = Kinetic Energy

In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.

The increase in tension on the steel wire is 8,484.75 N.

The given parameters;

The area of the steel wire is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (2\times 10^{-3})^2\\\\A = 1.257 \times 10^{-5} \ m^2[/tex]

The extension of the steel wire is calculated as follows;

[tex]\Delta l = \alpha \times l\times \Delta T\\\\\Delta l = (12\times 10^{-6}) \times (8) \times (10 + 273)\\\\\Delta l = 0.027 \ m[/tex]

The increase in tension on the steel wire is calculated as follows;

[tex]E = \frac{stress}{strain } = \frac{\ F/A}{\Delta l/l} \\\\E = \frac{F\times l}{A \times \Delta l} \\\\F = \frac{E\times A \times \Delta l }{l} \\\\F = \frac{(2\times 10^{11}) \times (1.257\times 10^{-5})\times 0.027}{8} \\\\F = 8,484.75 \ N[/tex]

Thus, the increase in tension on the steel wire is 8,484.75 N.

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60 inches

because if 30=50% then 60=100% so its 60 inches

The sum of the series allows to find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop   height    rebound

   1        30          15

   2       15            7.5

   3        7.5         3.75

If we call the first term y₀  

The first bounce can be found.

                             y₁ = [tex]\frac{y_o}{2}[/tex]

The second bounces.

                             [tex]y_2 = \frac{y_1}{2} \\y_2 = \frac{y_o}{4}[/tex]

The third bounce.

                             [tex]y_3 = \frac{y_2}{2} \\y_3 = \frac{y_0}{ 8}[/tex]

By observing this table we can construct a series of the form

      Total distance = [tex]y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+ \frac{1}{8} + ... +\frac{1}{2n} )[/tex]

The sum of the serie has a result of

        sum  = 127/64 = 1,984

Let's calculate

     distance total  = 30 1,984

     Distance total = 59.52 cm

In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

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Answer:

972 J

Explanation:

At the bottom, all the gravitational potential energy was converted into kinetic energy. If you calculate the GPE, its value will be the same that the KE at the bottom. The GPE can be calculated this way:

GPE = mass×gravity×heigth

GPE = 2.2×9.8×45.08 ≈ 972