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guapka [62]
3 years ago
8

a boy is standing 4 meter from a plane mirror how far and in what distance must te move so that he will be 4 meter from his imag

e​
Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

2 meters towards the mirror.

Explanation:

In a plane mirror the image distance is equal to the object distance. Therefore, by moving 2 meters towards the mirror, the boy reduces the distance between him and the mirror to two meters which is the object distance. The image distance is also 2 meters. add the two distances you will get four meters.

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Formular for coefficient of friction<br>​
Bumek [7]

Answer:

hope you like it mark as brainliest

Explanation:

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact.

The coefficient of friction (fr) is a number that is the ratio of the resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together. It is represented by the equation: fr = Fr/N.

7 0
3 years ago
Um aluno pretende construir um aquecedor usando um enrolamento de fio. Experimenta e verifica que não
Vlad1618 [11]

Answer:

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6 0
3 years ago
Block B (of mass m) is initially at rest. Block A (of mass 3m) travels toward B with an initial speed v0 (vee nought) and collid
NeTakaya

Answer:

The maximum height reached by the two blocks is approximately 0.1147959 × v₀²

Explanation:

The mass of block B = m

The mass of block A = 3·m

The initial velocity of block B, v₂ = 0 m/s

The initial velocity of block A, v₁ = v₀

The amount of friction between the blocks and the surface = Negligible friction

By the Law of conservation of linear momentum, we have;

Total initial momentum = Total final momentum

3·m·v₁ + m·v₂ = (3·m + m)·v₃ = 4·m·v₃

Plugging in the values for the velocities gives;

3·m × v₀ + m × 0 = (3·m + m)·v₃ = 4·m·v₃

∴ 3·m × v₀ =  4·m·v₃

\therefore v_3 = \dfrac{3}{4} \times v_0 = 0.75 \times v_0

The kinetic energy, K.E. of the combined blocks after the collision is given as follows;

K.E. = 1/2 × mass × v²

\therefore K.E. = \dfrac{1}{2} \times 4\cdot m \times \left (\dfrac{3}{4} \cdot v_0 \right )^2 = \dfrac{9}{8} \cdot m\cdot v_0^2

The potential energy, P.E., gained by the two blocks at maximum height = The kinetic energy, K.E., of the two blocks before moving vertically upwards

The potential energy, P.E. = m·g·h

Where;

m = The mass of the object at the given height

g = The acceleration due to gravity

h = The height at which the object of mass, 'm', is located

Therefore, for h = The maximum height reached by the two blocks, we have;

P.E. = K.E.

m \cdot g \cdot h =  \dfrac{9}{8} \cdot m\cdot v_0^2

h = \dfrac{\dfrac{9}{8} \cdot m\cdot v_0^2}{m \cdot g }  = \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ g }  =  \dfrac{9}{8} \cdot \dfrac{ v_0^2}{ 9.8} = \dfrac{42}{392} \cdot  v_0^2 \approx 0.1147959   \cdot  v_0^2

The maximum height reached by the two blocks, h ≈ 0.1147959·v₀².

7 0
3 years ago
Tim, with mass 53.5 kg, climbs a gymnasium
Stella [2.4K]
Potential energy = mgh

= 53.5 x 9.8 x 4.7

= 2464.21 Joules
7 0
3 years ago
a 2.0 kg ball is dropped from a height of 20 m onto a soft surface and rebounds to a height of 5.0 m . what is the magnitude of
Yuki888 [10]

Based on the data provided, the impulse of the floor on the ball is 59.4 Ns.

<h3>What is the impulse of the floor on the ball?</h3>

Using the equation of motion to determine the velocity at the end of the fall

  • v^2 = u^2 + 2gh

Where v is velocity at the end of fall

u is initial velocity = 0

g is acceleration due to gravity = 9.81 m/s^2

h is height = 20

  • Taking downward velocity as negative and up as positive

v^2 = 0 + 2 (9.81)(20)

v^2 = 392.4

v = - 19.8 m/s

The velocity, v after bouncing is calculated also:

u = 0

g = 9.81 m/s^2

h = 5.0 m

v^2 = 0 + 2(9.81)(5)

v^2 = 98.1

v = 9.904 m/s

  • Impulse = change in momentum
  • Impulse = m(v- u)

Impulse = 2.0 × (9.9 -(-19.8)

Impulse = 59.4 Ns

Therefore, the impulse of the floor on the ball is 59.4 Ns.

Learn more about impulse at: brainly.com/question/904448

5 0
2 years ago
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