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bogdanovich [222]
2 years ago
10

A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .

Physics
1 answer:
pickupchik [31]2 years ago
7 0

The thermal energy that is generated due to friction is 344J.

<h3>What is the thermal energy?</h3>

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 *  2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

Learn more about thermal energy due to friction:brainly.com/question/7207509

#SPJ1

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6 0
3 years ago
1. It takes a 25 N force to push a 250 N box across the floor.
Fed [463]

Answer:

A:- 50 J

B:- 500 J

Explanation:

a) Given that a 25 N force is applied to move the box. Also the floor is having friction surface.

So in order to move the box, the floor should have friction of atleast 25 N.

∴ friction = 25 N

Work done = force * displacement of box

Given, the box is moved 2 m across the floor

So, Work done = friction * 2 m

                         = 25 * 2

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b) Given, the box is having weight of 250 N weight

Gravitational force is acting on the box which is equal to (mass * gravity)

∴ Force = 250 N

The box is lifted 2 m above the floor.

So, displacement = 2 m

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5 0
3 years ago
Two 10-cm-diameter charged rings face each other, 25 cm apart. The left ring is charged to ? 25 nC and the right ring is charged
MArishka [77]

Answer:

A)   E = 0N/C

B)   0i + 0^^j

C)   F = 0N

D)   0^i  + 0^j

Explanation:

You assume that the rings are in the zy plane but in different positions.

Furthermore, you can consider that the origin of coordinates is at the midway between the rings.

A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.

You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:

E=k\frac{rQ}{(r+R^2)^{3/2}}               (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C

Q: charge of the ring

r: perpendicular distance to the center of the ring

R: radius of the ring

You use the equation (1) to calculate the net electric field at the midpoint between the rings:

E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.

B) The direction of the electric field is 0^i + 0^j

C) The magnitude of the force on a proton at the midpoint between the rings is:

F=qE=q(0N/C)=0N

D) The direction of the force is 0^i + 0^j

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Option (B)is correct.

Alternating electric and magnetic fields has electromagnetic radiation that travel in the form of a wave.

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