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Alecsey [184]
3 years ago
8

In unit-vector notation, what is the net torque about the origin on a flea located at coordinates (-2.0, 1.0 m, 1.0 m) when forc

es F1 = (1.0 N) k and F2 = (-2.0 N) j act on the flea?
Physics
1 answer:
djverab [1.8K]3 years ago
4 0

Answer:

Torque, \tau=3i+2j+4k

Explanation:

Given that,

Position of the flea, r=(-2i+j+k)\ m

Force acting on the flea, F=(-2j+k)\ N

We need to find the net torque about the origin on a flea located at coordinates. Its formula is given by :

\tau=r\times F

\tau=(-2i+j+k) \times (-2j+k)

On solving the cross product of r and F, we get :

\tau=3i+2j+4k

So, the net torque about the origin on a flea is 3i+2j+4k. Hence, this is the required solution.

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A mass of 30.0 grams hangs at rest from the lower end of a long vertical spring. You add different amounts of additional mass ΔM
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Answer:

K = 373.13 N/m

Explanation:

The force of the spring is equals to:

Fe - m*g = 0     =>    Fe = m*g

Using Hook's law:

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K = m/X * g

In this equation, m/X is the inverse of the given slope. So, using this value we can calculate the spring's constant:

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A skier is accelerating down a 30.0-degree hill at 3.80 m/s^2.
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Answer:

ax = -3.29[m/s²]

ay = -1.9[m/s²]

Explanation:

We must remember that acceleration is a vector and therefore has magnitude and direction.

In this case, it is accelerating downwards, therefore for a greater understanding we will make a diagram of said vector, this diagram is attached.

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A student pulls a 50-newton sled with a force having a magnitude of 15 newtons. What is the magnitude of the force that the sled
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Answer:

Force = 35 N

Explanation:

From Newton's third law of motion, the boy must apply a force greater than the weight of the sled to lift it.

weight of sled = mg

where m is its mass and g the force of gravity on it.

weight of sled = 50 N

Force applied by the boy on the sled = 15 N

Since the force applied on the sled by the boy is lesser than the weight of the sled, then;

Force that the sled exerts on the student = 50 - 15

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The force exerted by the sled on the student is 35 N.

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