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andreev551 [17]
3 years ago
8

Draw the force vector starting at the black dot. The location, orientation, and length of the vector will be graded. You can mov

e the vectors F⃗ 1F→1 and F⃗ 2F→2 to construct the required vector, but be sure to return them into their initial positions before submitting the answer.

Physics
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

F3 is the equilibrant force equal in magnitude to the resultant between F1 and F2 but opposite in direction to it.

Explanation:

Given the diagram, the force F3 to make the body remain at rest or in equilibrium is the equilibrant force, this force is equal in magnitude to the resultant Fr between F1 and F2 but opposite in direction to it.

See attachment for diagram of forces.

The resultant force;

Fr =√ (F1)² + (F2)²...(1) [diagram a]

Therefore the length of F3 is Fr

F3 = -Fr

The direction (diagram b) of the resultant force Fr is given by

∆ = acrtan[(F1/F2)]

The direction of F3 is [90 + arctan(F1/F2)]

As seen in the diagram (d), the location of the force is in the fourth quadrant.

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10

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3 years ago
In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s
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a) 0.94 m

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W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where:

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d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

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b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

I=\Delta p = m \Delta v

where

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I=(68 kg)(-55 m/s)=-3740 kg m/s

7 0
3 years ago
A spring is stretched from x=0 to x=d, where x=0 is the equilibrium position of the spring. It is then compressed from x=0 to x=
VARVARA [1.3K]

Answer:

The same amount of energy is required to either stretch or compress the spring.

Explanation:

The amount of energy required to stretch or compress a spring is equal to the elastic potential energy stored by the spring:

U=\frac{1}{2}k (\Delta x)^2

where

k is the spring constant

\Delta x is the stretch/compression of the spring

In the first case, the spring is stretched from x=0 to x=d, so

\Delta x = d-0=d

and the amount of energy required is

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In the second case, the spring is compressed from x=0 to x=-d, so

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and the amount of energy required is

U=\frac{1}{2}k (-d)^2= \frac{1}{2}kd^2

so we see that the amount of energy required is the same.

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