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Snowcat [4.5K]
3 years ago
11

Can someone pls help me

Physics
2 answers:
andriy [413]3 years ago
7 0
B hope this helps ._.
natali 33 [55]3 years ago
5 0

Answer:

if not mistaken its b

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What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving
lions [1.4K]

Complete question is;

Jason works for a moving company. A 75 kg wooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11°.

What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving UP the ramp?

Answer:

F = 501.5 N

Explanation:

We are given;

Mass of wooden crate; m = 75 kg

Angle of ramp; θ = 11°

Now, for the wooden crate to slide upwards, it means that the force of friction would be acting in an opposite to the slide along the inclined plane. Thus, the force will be given by;

F = mgsin θ + μmg cos θ

From online values, coefficient of friction between wooden surfaces is μ = 0.5

Thus;

F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)

F = 501.5 N

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3 years ago
To test the quality of a tennis ball, you drop it onto the floor from a height of 4.00 m. It rebounds to a height of 2.00 m. If
arlik [135]

Answer:

Part a)

a = 1260.3 m/s^2

Part b)

Direction = upwards

Explanation:

When ball is dropped from height h = 4.0 m

then the speed of the ball just before it will strike the ground is given as

v_f^2 - v_i^2 = 2 a d

v_1^2 - 0^2 = 2(9.81)(4.0)

v_1 = 8.86 m/s

Now ball will rebound to height h = 2.00 m

so the velocity of ball just after it will rebound is given as

v_f^2 - v_i^2 = 2 a d

0 - v_2^2 = 2(-9.81)(2.00)

v_2 = 6.26 m/s

Part a)

Average acceleration is given as

a = \frac{v_f - v_i}{\Delta t}

a = \frac{6.26 - (-8.86)}{12.0 \times 10^{-3}}

a = 1260.35 m/s^2

Part B)

As we know that ball rebounds upwards after collision while before collision it is moving downwards

So the direction of the acceleration is vertically upwards

7 0
3 years ago
A basketball is shot from 2 meters up at an angle of 60° above the x axis at an initial velocity of 9 m/s. What is the maximum h
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Hope this helps a little

initial distance up = 2

initial velocity component up = 9 sin 60 = 7.79

v = 9 sin 60 - 9.8 t

when v = 0, we are there

9.8 t = 7.79

t = .795 seconds to top

h = 2 + 7.79(.795) - 4.9(.795^2)

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3 years ago
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How can an animal regeneration produce two results?
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The unit for measuring electric power is the
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