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Tamiku [17]
3 years ago
14

True or false conceptual physics 2 questions from chapter 25.

Physics
1 answer:
Arlecino [84]3 years ago
3 0

Answer:

Explanation:

1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.

2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.

3) TRUE, W = q.VBA.

4) FALSE; eV is a unit for work (or) energy.

5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.

6) FALSE; in the direction of electric field the potential decreases.

7) FALSE; equipotential surface is perpendicular to the electric field lines.

8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.

9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.

10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.

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A 1100 kg car rounds a curve of radius 68 m banked at an angle of 16 degrees. If the car is traveling at 95 km/h, will a frictio
Mariulka [41]

Answer:

Yes. Towards the center. 8210 N.

Explanation:

Let's first investigate the free-body diagram of the car. The weight of the car has two components: x-direction: towards the center of the curve and y-direction: towards the ground. Note that the ground is not perpendicular to the surface of the Earth is inclined 16 degrees.

In order to find whether the car slides off the road, we should use Newton's Second Law in the direction of x: F = ma.

The net force is equal to F = \frac{mv^2}{R} = \frac{1100\times (26.3)^2}{68} = 1.1\times 10^4~N

Note that 95 km/h is equal to 26.3 m/s.

This is the centripetal force and equal to the x-component of the applied force.

F = mg\sin(16) = 1100(9.8)\sin(16) = 2.97\times10^3

As can be seen from above, the two forces are not equal to each other. This means that a friction force is needed towards the center of the curve.

The amount of the friction force should be 8.21\times 10^3~N

Qualitatively, on a banked curve, a car is thrown off the road if it is moving fast. However, if the road has enough friction, then the car stays on the road and move safely. Since the car intends to slide off the road, then the static friction between the tires and the road must be towards the center in order to keep the car in the road.

5 0
4 years ago
If continuous spectrum light from a star passes through a cool, low-density gas on its way to your telescope and spectroscope, _
natima [27]

Answer: a dark (absorption) line

Explanation:

This is as a result of absorption of electromagnetic a specific wavelength. The pattern followed by such lines is characteristic of specific atoms in the path of radiation.

6 0
3 years ago
I really need help with this!! Will give brainliest if you help T^T
slava [35]

Answer:

What inferences can you make about the melting points of the different substances and the motion of their particles, based on the data? (ignore that needed it here so i could see it better.)

Explanation:

Butter has a lower melting point than the cheese and the wax. The motion of the cheese were a little separated while the butter articles have more space in between. The wax had the closest particles.

I dont know if that makes sense? 

5 0
3 years ago
Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w
musickatia [10]
The net force on the box is:
50 + 30 - 65
= 15 Newtons
This can be used in conjunction with
F = ma
to calculate the acceleration of the box if its mass is known.
3 0
4 years ago
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.5 kg⋅m2 and for arms and legs in is 0.80
sergiy2304 [10]

Answer:

1.3  rev/s

Explanation:

I_{o} = Moment of inertia when arms and one leg of the skater is out = 3.5 kgm²

I_{i} = Moment of inertia when arms and  legs of the skater are in = 0.80 kgm²

w_{i} = Angular speed of skater when arms and  legs of the skater are in = 5.5 rev/s

w_{o} = Angular speed of skater when arms and  legs of the skater are out = ?

Using conservation of angular momentum

I_{i} w_{i} = I_{o} w_{o}

(0.80) (5.5) = (3.5) w_{o}

w_{o} = 1.3 rev/s

8 0
3 years ago
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