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Mrac [35]
3 years ago
10

In which one of the following situations is zero net work done? A) A ball rolls down an inclined plane. B) A physics student str

etches a spring. C) A projectile falls toward the surface of Earth. D) A box is pulled across a rough floor at constant velocity. E) A child pulls a toy across a rough surface causing it to accelerate.
Physics
2 answers:
spayn [35]3 years ago
8 0

Answer:

D

Explanation:

I have taken the test and yeah love ya

lidiya [134]3 years ago
6 0

Answer:

Option D

Explanation:

The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.

In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.  

In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.

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During which stage does the birth rate begin to decline?
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Answer:

During stage 3 - late expanding (of demogrpahic transition model)

Explanation:

During stage 3, birth rate begins to decline as infant mortality is lower and women have more access to education, family planning, and contraceptives.  Children are not needed as "free labor" as they might have been in earlier stages.

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3 years ago
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The nucleus contains protons and neutrons. Being positively charged, the protons repel each other. The nucleus should fly apart
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this is due to the existence of other forces called the strong nuclear forces that overcomes the repulsion forces between the protons and keeps the nucleons holding to each other also there is a type of energy that is called the nuclear binding energy and this energy also works on binding the components of the nucleus together

6 0
3 years ago
Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.
Vitek1552 [10]

Answer:

Explanation:

Explain how a projectile might be modified to decrease the air resistance impacting its trajectory.

5 0
2 years ago
A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1
pav-90 [236]
I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
F_a=F_cf-F_g
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
F_a=m\frac{v^2}{r}-mg=179 $N
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
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6 0
3 years ago
Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h
mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(8.89² - 6.11²)

                                                    = 20.85m J

 

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(11.67² - 8.89²)

                                                    = 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

6 0
3 years ago
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