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lina2011 [118]
3 years ago
6

Suppose a rock is hoisted up with a rope. The rock experiences 10 N of tension force upward and 10 N of gravitational force down

ward. The total net force on the rock is 20 N. True False
Physics
2 answers:
nika2105 [10]3 years ago
5 0
False. Since the forces are pulling in equal and opposite directions, the net force is 0.
LiRa [457]3 years ago
5 0

10N of tension upwards is <em>positive</em>

10N of gravitational Force downwards is <em>negative</em>

Net Force = The sum of all forces

F_{net} = F_{tension} + F_{gravity}

F_{net} = 10N + (-10N) = 0


So the answer is false.

Extra Info: If the net force was 20N there would be a tensional force of 30N moving the rock upwards.

30N - 10N = 20N

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The Force on the left hand pole, F' = 0.167N

<h3>What is the force on the left hand pole?</h3>

Force is an agent which produces a change in the motion or state of an object.

Force is a vector quantity.

The general force is calculated as follows:

F = mg/sinθ

m = 17.1 g = 0.0171 kg

g = 9.81 m/s²

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F = 0.0171 * 9.81/sin45

F = 0.237 N

Force on the left hand pole, F' = Fcosθ

F' = 0.237 * cos 45

F' = 0.167N

In conclusion, the force on the left hand pole is the horizontal component of force.

Learn more about force at: brainly.com/question/141439

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In an elastic collision, a 400-kg bumper car collides directly from behind with a second, identical bumper car that is traveling
kirza4 [7]

Answer:

v₁ = 3.5 m/s

v₂ = 6.4 m/s

Explanation:

We have the following data:

m₁ = mass of trailing car = 400 kg

m₂ = mass of leading car = 400 kg

u₁ = initial speed of trailing car = 6.4 m/s

u₂ = initial speed of leading car = 3.5 m/s

v₁ = final speed of trailing car = ?

v₂ = final speed of leading car = ?

The final speed of the leading car is given by the following formula:

v_2=\frac{2m_1}{m_1+m_2}u_1-\frac{m_1-m_2}{m_1+m_2}u_2\\\\v_2=\frac{(2)(400\ kg)}{400\ kg+400\ kg}(6.4\ m/s)-\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(3.5\ m/s)

<u>v₂ = 6.4 m/s</u>

The final speed of the leading car is given by the following formula:

v_1=\frac{m_1-m_2}{m_1+m_2}u_1+\frac{2m_2}{m_1+m_2}u_2\\\\v_1=\frac{400\ kg-400\ kg}{400\ kg + 400\ kg}(6.4\ m/s)+\frac{(2)(400\ kg)}{400\ kg+400\ kg}(3.5\ m/s)

<u>v₁ = 3.5 m/s</u>

4 0
3 years ago
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