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AURORKA [14]
3 years ago
6

A man is walking away from a lamppost with a light source h = 6 m above the ground. the man is m = 2 m tall. how long is the man

's shadow when he is d = 8 m from the lamppost?
Physics
1 answer:
Llana [10]3 years ago
3 0

Answer;

= 4 m is the length of the man's shadow.

Explanation;

2/x=6/(8+x) cross multiply.

6x=2(8+x)

6x=16+2x

6x-2x=16

4x=16

x=2=16/4

x=4 m. is the length of the man's shadow.

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A child whose weight is 235 N slides down a 4.90 m playground slide that makes an angle of 37.0° with the horizontal. The coeffi
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Answer:

Explanation:

a) First, let's calculate the value of the Friction force, which is given by the formula:

Ff = u*W

As the Friction force has an X component, it would be:

Ff = u*m*g*cosФ

Where m*g is the weight of the child.

Solving for Ff:

Ff = 0.051 * 235cos37 = 9.57 N

Now, to get the energy transferred to thermal energy (or heat) we need to get the Work done, so:

Wf = Ff * d

Wf = 9.57 * 4.9 = 46.9 J

b) We need to get the downslope component of the child weight, which is:

Wy = 235sin37 = 141.43 N

As you can imagine, the gravity also does work, so:

Wg = Wy * d

Wg = 141.43 * 4.9 = 693 J

Now, let's get the kinetic energy, which can be obtained by this expression:

ΔKe = Wg - Wf

ΔKe = 693 - 46.9 = 646.1 J

The formula for kinetic energy is:

Ke = 1/2 m*V^2

We have the innitial speed which is 0.355 m/s, and the mass can be obtained by m*g so:

Fw= m*g ----> m = Fw/g

m = 235 / 9.8 = 23.97 kg

so the innitial energy is:

Ke = 1/2 * 23.97 * (0.355)^2

Ke = 1.51 J

This could be Ke1, so to get Ke2:

ΔKe = Ke2 - Ke1

Ke2 = Δke + Ke1

Ke2 = 646.1 + 1.51 = 647,61 J

Finally, the speed at the bottom would be:

v = √2Ke/m

v = √2*647.61/23.97

v = 7.35 m/s

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Answer:

1st statement is true

Explanation:

Here statement 1 is correct

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5 0
4 years ago
A student pulls a sled of mass m = 82.0 kg with a force of F = 160N, and the force makes an angle of θ = 15 degrees with respect
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Answer:B

Explanation:

Given

mass of sledge=82 kg

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8 0
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answer : P = 320 W

........

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