A runner starts at position zero and moves in one direction to position 12m 3. What is the distance traveled by the runner?

Physics

1 answer:

murzikaleks [220]2 years ago

5 0

Distance = 12 m

Explanation:

Distance covered by the runner is same as the total path that he has ran. Here, a person starts at 0 position and goes in one direction to position 12 m. Since the initial point is 0 and the final position is 12, the total distance is calculated by adding the 2 points as: 0+12 = 12 m. So the distance covered and the displacement both are same which is 12 meters.

Distance traveled is the full length of the path covered between two points. It is not a vector and the direction is nil and no negative sign. Example : The distance walked by the teacher is 3.0 m ,The distance walked by a passenger is 5.0 m. Also, the distance traveled need not be equal to the magnitude of the displacement which is the distance between the two positions.

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A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary

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The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

mass of the bullet, m = 23 g = 0.023 g
speed of the bullet, u = 230 m/s
mass of the wood, m = 2 kg
final speed of the bullet, v = 170 m/s
coefficient of friction, μ = 0.15

The final velocity of the wood after the bullet hits is calculated as follows;

$m_1u_1 + m_2 u_2 = m_1v_1 + m_2v_2\\\\0.023(230) + 2(0) = 0.023(170) + 2v_2\\\\5.29 = 3.91 + 2v_2\\\\2v_2 = 1.38\\\\v_2 = \frac{1.38}{2} = 0.69 \ m/s$

The acceleration of the wood is calculated as follows;

$\mu = \frac{a}{g} \\\\a = \mu g\\\\a = 0.15 \times 9.8\\\\a = 1.47 \ m/s^2$

The distance traveled by the wood after the bullet emerges is calculated as follows;

$v^2 = v_0^2 + 2as\\\\v^2 = 0 + 2as\\\\v^2 = 2as\\\\s = \frac{v^2}{2a} \\\\s = \frac{(0.69)^2}{2(1.47)} \\\\s = 0.16 \ m$

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.

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3 years ago

Read 2 more answers

(15 Points)

oksian1 [2.3K]

The vertical weight carried by the builder at the rear end is F = 308.1 N

<h3>Calculations and Parameters</h3>

Given that:

The weight is carried up along the plane in rotational equilibrium condition

The torque equilibrium condition can be used to solve

We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person

This would lead to:

F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)

F(1cos20)= 197/2(3.10sin20 + 2 cos 20)

Fcos20= 289.55

F= 308.1N

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1 year ago

Please help I will mark you brainliest

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