The charge 0.00068 C can be written using the prefix µ as 680 μC.
How to eplain how the charge is to be writtenThe prefix "µ" (pronounced "micro") represents a scaling factor of 10⁻⁶ . This means that when you write a value with the prefix µ, it is equal to the value multiplied by 10⁻⁶. I
In this case, 680 μC means 680 times 10⁻⁶ coulombs, which is equal to 0.00068 C.
So, the correct way to express the charge 0.00068 C using the prefix µ is 680 μC.
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400 kg cart with rubber wheel just stopped in a while asphalt road because it ran out of gas attractive attaches a route to the golf cart to tow it in Post Achurr down the road accelerating the golf cart at three Emperor a Square what is the tension in the rope
To solve this problem, we will use Newton's second law of motion.
Newton's second law states that the force is equal to mass times acceleration.
Hence, the formula is as follows:Force = mass x acceleration
To calculate the force, we need to know the mass of the cart and the acceleration at which it is being towed.
As per the question, the cart's mass is 400 kg, and it is being accelerated at a rate of 3 m/s^2.
Hence, the force required to accelerate the cart can be calculated as follows:
Force = mass x acceleration [tex]Force = 400 kg\times3 m/s^2[/tex]
[tex]Force = 1200 N[/tex]
Therefore, the tension in the rope is 1200 N.
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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?
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.For such more question on velocity
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Exercise 3 Read the following statement: If I switch from using incandescent light bulbs to LED light bulbs, then my electricity bill will decrease. In this situation, the type of light bulb used is the MENU variable
If I switch from using incandescent light bulbs to LED light bulbs, then my electricity bill will decrease. In this situation, the type of light bulb used is the Independent variable.
What is the independent variable?The independent variable in a circumstance refers to the element that is not affected by the other variables.
In this case, the switching from incandescent lighting to LED bulbs causes a decrease in the electricity bill. So, the light bulb is the independent variable that is unaffected by the other factors.
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why is nuclear fusion not used to produce electricity
Answer:
it is incredibly difficult to sustain a fusion reaction.
Explanation:
The main reason we aren't using nuclear fusion to generate power yet is because it is incredibly difficult to sustain a fusion reaction. The energy requirements are very high, and it is hard to find materials that can withstand such high temperatures.
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?
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 loseTo 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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what are crystalline substances in physics
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 unique strengths would you bring to the Grace Scholars Program?
Grace Scholars Program is a scholarship program that selects the best and brightest students for the program.
The Grace Scholars Program is a prestigious scholarship program that aims to identify and support exceptionally talented and promising students. The program seeks to provide opportunities for these students to excel academically, develop their leadership skills, and make a positive impact in their respective fields.The selection process for the Grace Scholars Program is highly competitive, with a rigorous evaluation of applicants' academic achievements, extracurricular activities, personal qualities, and potential for future success. The program typically looks for students who demonstrate outstanding academic performance, intellectual curiosity, leadership abilities, and a commitment to service and community involvement. By offering this scholarship program, institutions aim to attract and retain top talent, foster a culture of excellence, and contribute to the development of future leaders and innovators who can positively impact society.For such more questions on scholarship
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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?
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.
What's elastic energy simple meaning please
Answer:
Elastic potential energy is energy stored as a result of applying a force to deform an elastic object.
on which factor does the mass of objects depend?
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 stone falls From rest From the Top of a Higher tower. Ignore Air resistance and take g=9.8m/s2. calculate:
a. speed of the stone after 2 seconds
b. how far the stone has travelled after 2 seconds
a)It takes 0 seconds for the stone to hit the ground.
b)The stone has traveled 19.6 meters after 2 seconds.
When an object is dropped from a height, the motion is described as free fall. It is a special form of motion in which an object is under the sole influence of gravity. The gravitational force acting on an object is always directed towards the center of the Earth, i.e., downwards. Therefore, when an object is dropped from a height, it falls freely towards the ground, and the acceleration of the object is equal to the acceleration due to gravity, which is approximately 9.8 m/s².
In the problem given, a stone is dropped from rest from the top of a higher tower. Therefore, the initial velocity of the stone, u = 0 m/s.
a. Time taken by the stone to reach the ground:
We can use the following equation of motion to calculate the time taken by the stone to reach the ground.
v = u + at
where,
v = final velocity of the object
u = initial velocity of the object
a = acceleration of the object
t = time taken by the object to reach from initial velocity to final velocity
Since the stone is dropped from rest, the initial velocity of the stone, u = 0 m/s.
At the ground level, the final velocity of the stone, v = ?
Again, the acceleration due to gravity, a = 9.8 m/s².
Therefore, we can write the equation as:
v = u + at
v = 0 + 9.8×t
v = 9.8t m/s
When the stone hits the ground, the final velocity of the stone, v = 0 m/s.
Therefore, we can rewrite the equation as:
0 = 9.8t
t = 0 seconds (when the stone is at the top of the tower)
We have found that it takes 0 seconds for the stone to hit the ground.
b. How far the stone has travelled after 2 seconds:
We can use the following equation of motion to calculate how far the stone has travelled after 2 seconds.
s = ut + (1/2)at²
where,
s = displacement of the object
u = initial velocity of the object
a = acceleration of the object
t = time taken by the object to reach from initial velocity to final velocity
Since the stone is dropped from rest, the initial velocity of the stone, u = 0 m/s.
The acceleration due to gravity, a = 9.8 m/s².
Therefore, we can write the equation as:
s = ut + (1/2)at²
s = 0×2 + (1/2)×9.8×(2)²
s = 19.6 m
Therefore, the stone has traveled 19.6 meters after 2 seconds.
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I need the answer to question 15 only, please
Answer:
The answer is C
Explanation:
My sister took that before on paper and she got a 78%
Pls helppp
Bumper car A (281 kg) moving +2.82 m/s
makes an elastic collision with bumper
car B (265 kg) moving -1.72 m/s. What is
the velocity of car A after the collision?
V
(Unit = m/s)
Remember: right is +, left is -
Enter
Explanation: For some reason, it would not let me click the "ADD YOUR ANWER" button, because it stated that my answer had some sort of "Inappropriate" word or something, so instead I attached two images of the answer that I had written down.
A soft tennis ball is dropped onto a hard floor from a height of 1.60 m and rebounds to a height of 1.28 m. (Assume that the positive direction is upward.)
(a)Calculate its velocity (in m/s) just before it strikes the floor. (answer is -5.6)
(b)Calculate its velocity (in m/s) just after it leaves the floor on its way back up. (answer is 5.01)
(c)Calculate its acceleration (in m/s2) during contact with the floor if that contact lasts 3.50 ms. (answer is 3031)
(d)How much (in m) did the ball compress during its collision with the floor, assuming the floor is absolutely rigid? ( I don't know this one)
(a) The velocity of the tennis ball just before it strikes the floor is -5.6 m/s.
(b) The velocity of the tennis ball just after it leaves the floor on its way back up is 5.01 m/s.
(c) The acceleration of the tennis ball during contact with the floor, assuming the contact lasts 3.50 ms, is 3031 [tex]m/s^2[/tex].
(d) The amount of compression the ball experienced during its collision with the floor, assuming the floor is absolutely rigid, is unknown.
(a) To calculate the velocity of the tennis ball just before it strikes the floor, we can use the principle of conservation of energy. The potential energy at the initial height is converted into kinetic energy just before it hits the floor.
Using the formula for gravitational potential energy (PE = mgh) and kinetic energy (KE = 0.5[tex]mv^2[/tex]), we can equate the two energies:
mgh = 0.5[tex]mv^2[/tex]
Here, m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 [tex]m/s^2[/tex]), h is the initial height (1.60 m), and v is the velocity just before it strikes the floor.
Simplifying the equation, we get:
9.8 * 1.60 = 0.5 * [tex]v^2[/tex]
15.68 = 0.5 * [tex]v^2[/tex]
31.36 = [tex]v^2[/tex]
Taking the square root of both sides, we find:
v = ±√31.36
Since the positive direction is upward, the velocity just before it strikes the floor is -5.6 m/s.
(b) To calculate the velocity just after the ball leaves the floor on its way back up, we can use the principle of conservation of energy again. The potential energy at the lowest point (the height of the ball during contact with the floor) is converted into kinetic energy.
Using the same equation as before, but with the final height of 1.28 m, we have:
9.8 * 1.28 = 0.5 * [tex]v^2[/tex]
12.544 = 0.5 * [tex]v^2[/tex]
25.088 = [tex]v^2[/tex]
v = ±√25.088
Since the positive direction is upward, the velocity just after it leaves the floor is 5.01 m/s.
(c) To calculate the acceleration during contact with the floor, we can use the formula for acceleration:
a = (v_f - v_i) / t
Here, v_f is the final velocity (5.01 m/s), v_i is the initial velocity (-5.6 m/s), and t is the duration of contact (3.50 ms = 0.0035 s).
a = (5.01 - (-5.6)) / 0.0035
a = 10.61 / 0.0035
a ≈ 3031 [tex]m/s^2[/tex]
Therefore, the acceleration during contact with the floor is approximately 3031 [tex]m/s^2[/tex].
(d) The amount of compression the ball experiences during its collision with the floor can be calculated using the principles of impulse and momentum.
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An electric iron is Mark 120 volts and 500 Watts to units consumed by it in using it for 24 hours will be
An electric iron is marked 120 volts and 500 Watts. The units consumed by it in using it for 24 hours can be calculated using the formula:Power (in watts) = Voltage (in volts) x Current (in amperes)P = V x I
Using the above formula, we can find the current drawn by the electric iron as follows:I = P/VI = 500/120I = 4.17 ATherefore, the power consumed by the electric iron in 24 hours is:P = VI x tP = 120 x 4.17 x 24P = 120 x 100.08P = 12010.56 watt-hoursTo convert watt-hours to kilowatt-hours, we divide by 1000: Energy consumed = 12010.56 / 1000Energy consumed = 12.01 kWhHence, the units consumed by the electric iron in 24 hours is 12.01 kilowatt-hours.For such more question on Voltage
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Which question asks for an opinion?
Answer:
blud forgot the photo
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?
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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if a 1500 kg car speeds up to 3.0 m/s and the gain is kinetic energy
Hello!
K.E. = 1/2mv²
m = mass
v = square
K.E.
= 1/2mv²
= 1/2 * 1500 * 3²
= 6750J
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
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.
, Not Selected
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.
Answer:
Number one would be a correct action-reaction pair.
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 -
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.For such more question on velocity
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Mantle fluid that is heating up will become ________ and move _________.
A. less dense, upwards
B. denser, upwards
C. less dense, downwards
D. denser, downwards
At orbital velocity around the Earth, the gamma factor is 1.000000000338. Russian cosmonaut Valery Polyakov spent 438 consecutive days orbiting the Earth on the Mir Space station. If he left a stopwatch running on Earth and took an identical stopwatch running with him on Mir, what would be the difference in elapsed time over the duration of his flight.
Answer Choices,
120 s
3.0 s
1.3 E−2 s
2.5 E−1 s
Answer:
The time dilation effect can be calculated using the time dilation formula, which is derived from the theory of special relativity. The formula is: Δt' = Δt / √(1 - v²/c²), where Δt is the time interval measured by a stationary observer, Δt' is the time interval measured by an observer in motion, v is the relative velocity between the two observers, and c is the speed of light.
In this case, Valery Polyakov spent 438 consecutive days orbiting the Earth on the Mir Space station at an orbital velocity where the gamma factor is 1.000000000338. The gamma factor is equal to 1 / √(1 - v²/c²), so we can solve for v²/c² = 1 - (1 / gamma)² = 2.28 x 10⁻¹⁸. Plugging this into the time dilation formula, we get that Δt' = Δt / √(1 - 2.28 x 10⁻¹⁸) ≈ Δt (1 + 1.14 x 10⁻¹⁸).
The time interval measured by a stationary observer on Earth is 438 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute = 37,843,200 seconds. Plugging this into the formula above, we get that Δt' ≈ 37,843,200 seconds (1 + 1.14 x 10⁻¹⁸) ≈ 37,843,200.000043 seconds.
Therefore, the difference in elapsed time between the stopwatch on Earth and the stopwatch on Mir over the duration of Valery Polyakov's flight would be approximately 0.000043 seconds or **4.3 x 10⁻⁵ seconds**. This answer is not among the answer choices you provided.
Answer:
1.3 E-2 s
Explanation:
Δt' = 438 days * 1.000000000338
Δt' = 438.00000000044484 days
Therefore, the difference in elapsed time between the stopwatch on Earth and the stopwatch on the Mir Space station over the duration of Valery Polyakov's flight would be approximately 0.00000000044484 days (or about 38.4 microseconds). 0.0384 seconds is closest to "1.3 E−2 s"
hi what's magnetic energy give SIMP LE answer please
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.
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)
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! =)
State three examples where undesirable on a
machine
In the field of machinery, various problems and challenges may arise during the operation of a machine. Three examples of undesirable conditions on a machine are overheating, excessive vibrations, and noise.
Overheating: When a machine is continuously working, it generates heat due to the resistance caused by the parts involved. However, if the machine becomes too hot, it may lead to severe damage to the machinery or the operators working with the machine. Overheating may occur due to several reasons such as lack of lubrication, excessive friction, or improper functioning of cooling systems.
Excessive vibrations: If a machine vibrates too much, it may cause the machine to loosen and break apart. Moreover, excessive vibrations can cause discomfort to the operator working with the machinery. Excessive vibration may arise due to improper balance, lack of maintenance, or wear and tear of the machine parts.
Noise: Machines produce sounds during their operations, but if the sound is too loud and continuous, it may cause damage to the operator's eardrums or cause distraction to people working around the machine. Noise pollution may occur due to unbalanced machine parts, lack of lubrication, and improper functioning of the machine's internal systems.These undesirable conditions on a machine need to be addressed to avoid severe damage to the machine or harm to operators working with the machinery. Regular maintenance, lubrication, and inspection of the machinery can minimize or eliminate the chances of these undesirable conditions on the machine.
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What is the change in internal energy if 70J of heat is added to a system and the system does 30J of work on the surroundings
Answer: Δ 100 J
Explanation:
Since, Q = 70 J (heat added to the system) and W = -30 J (work done by the system on the surroundings), we;
Substitute these values into the equation, we have:
ΔU = 70 J - (-30 J)
ΔU = 70 J + 30 J
ΔU = 100 J
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
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
Does ISO sensibilty in photography have a unit of measurement?
Ex: like lux, or nit, etc...
ISO sensitivity in photography does not have a specific unit of measurement.
ISO sensitivity in photography does not have a specific unit of measurement. ISO stands for International Organization for Standardization, and it is a standardized scale used to measure the sensitivity of a camera's image sensor to light. It is commonly referred to as "ISO speed" or "ISO value."
The ISO scale typically ranges from low values, such as ISO 100 or 200, to higher values like ISO 1600 or even higher. The higher the ISO value, the more sensitive the camera's sensor is to light, allowing for capturing images in low-light conditions without the need for longer exposure times.
ISO sensitivity is not expressed in units like lux or nit, which are measurements of illuminance or luminance, respectively. Lux measures the amount of light per unit area, while nit measures the brightness of a light source.
ISO sensitivity is a relative scale that represents the amplification of the sensor's signal. As the ISO value increases, the sensor's sensitivity is increased, but at the cost of introducing more digital noise and reducing image quality.
In summary, ISO sensitivity in photography does not have a specific unit of measurement. It is a standardized scale used to indicate the sensitivity of a camera's image sensor to light, allowing photographers to adjust exposure settings and capture images in various lighting conditions.
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A car mass 1200 kg is driven around a corner of radius 45m at 15 ms
Calculate the acceleration of the car
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.