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2 years ago

Please someone help me.

Physics

1 answer:

iren2701 [21]2 years ago

7 0

Answer:

1) The block of wood is at rest.

2) Weight and Normal Reaction (see attached diagram).

Explanation:

Assumptions

The block of wood is a particle (a body whose mass acts at a point, so its dimensions don't matter).
There are no other external forces involved.

Question 1

"At rest" or "resting" means that the object isn't moving.

Therefore, the block of wood is at rest.

Question 2

Forces that are acting on the block of wood:

Weight (W): Due to the particle's mass, m, and the acceleration due to gravity, g: W = mg
(Weight always acts downwards).

Normal Reaction (R or N): The reaction from a surface.
(Normal Reaction is always at 90° to the surface).

See attached diagram.

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Name three ways carbon and oxygen get into the environment.

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Humans,trees,plants hope it helps

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4 years ago

During a firework display, a shell is shot through the air with an initiak speed of 70 m/s at an angle of 75° above the horizont

ladessa [460]

Answer:

241.24m

Explanation:

The height at which the shell explodes will be at the maximum height. In projectile motion, maximum height formula is expressed as:

H = u²sin²θ/2g

u is the initial speed = 70m/s

θ the angle of launch = 75°

g is the acceleration due to gravity = 9.81m/s²

Substitute the values into the formula and get H

H = 70²(sin75°)/2(9.81)

H = 4900sin75°/19.62

H = 4900*0.9659/19.62

H = 4733.037/19.62

H = 241.24m

Hence the height at which the shell explodes is 241.24m

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3 years ago

The diagram below shows the oscillation of a

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Answer:

I think option B 0.125

Explanation:

but i am not sure so don't mind

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3 years ago

Water drops fall from the edge of a roof at a steady rate. a fifth drop starts to fall just as the first drop hits the ground. a

Alchen [17]

The height of the roof is 3.57m

Let the drops fall at a rate of 1 drop per t seconds. The first drop takes 5t seconds to reach the ground. The second drop takes 4t seconds to reach the bottom of the 1.00 m window, while the 3rd drop takes 3t s to reach the top of the window.

Calculate the distances traveled by the second and the third drops s₂ and s₃, which start from rest from the roof of the building.

$s_2=\frac{1}{2} g(4t)^2=8gt^2\\ s_3=\frac{1}{2} g(3t)^2=(4.5)gt^2$

The length of the window s is given by,

$s=s_2-s_3\\ (1.00 m)=8gt^2-4.5gt^2=3.5gt^2\\ t^2=\frac{1.00 m}{(3.5)(9.8m/s^2)} =0.02915s^2$

The first drop is at the bottom and it takes 5t seconds to reach down.

The height of the roof h is the distance traveled by the first drop and is given by,

$h=\frac{1}{2} g(5t)^2=\frac{25t^2}{2g} =\frac{25(0.02915s^2)}{2(9.8m/s^2)} =3.57 m$

the height of the roof is 3.57 m

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4 years ago

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3 years ago