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attashe74 [19]
4 years ago
11

The structures shown are primarily involved in

Physics
2 answers:
VLD [36.1K]4 years ago
6 0
Removing waste.
_____
Gennadij [26K]4 years ago
6 0
The correct answer is 3 (removing wastes)
Good luck with future problems m8
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A baseball player hits a ball with a bat. At one moment the force exerted by the bat on the ball is 18,000 N. At that same momen
11111nata11111 [884]

Answer:

Less than 18000N

Explanation:

Given

Force\ Exerted = 18000N

This question will be answered using Newton's third law.

Understanding this law, it implies that reaction force is equal and opposite to the force exerted.

This implies that;

If the force exerted on the ball is 18000N

the force exerted is -18000N

So, the option that answers the question is less than 18000N because -18000N < 18000N

5 0
3 years ago
What does the vibrating object do to the air particles around it ?
Westkost [7]

Answer:

When an object vibrates, it causes movement in surrounding air molecules. These molecules bump into the molecules close to them, causing them to vibrate as well. This makes them bump into more nearby air molecules.

7 0
3 years ago
How much force is needed to accelerate an object of mass 90 kg at a rate of 1.2 m/s2? 0.013 N 75 N 108 N 1080 N
lana [24]
F = m*a

F = 90*1,2

F = 108 N
8 0
4 years ago
Read 2 more answers
Why do cats have tails?
tester [92]
Cats have tails to help their balance. Similar to the stick a trapeze/high wire walker uses.

The tail helps to serve as a counterbalance when cats walk on narrow spaces such as fences or shelves. The tail also aids in balance when a cat is running after or jumping on prey.

Just so you know....Cats can live without tails.
3 0
3 years ago
. A projectile is fired with an initial velocity of 113 m słatan angle of 60.0
bonufazy [111]

Answer:

a) 9.99 s

b) 538 m

c) 20.5 s

d) 1160 m

Explanation:

Given:

x₀ = 0 m

y₀ = 49.0 m

v₀ = 113 m/s

θ = 60.0°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find t.

vᵧ = aᵧ t + v₀ᵧ

(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)

t ≈ 9.99 s

b) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)

y ≈ 538 m

c) When the projectile lands, y = 0 m.  Find t.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²

You'll need to solve using quadratic formula:

t ≈ -0.489, 20.5

Since negative time doesn't apply here, t ≈ 20.5 s.

d) When the projectile lands, y = 0 m.  Find x. (Use answer from part c).

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²

x ≈ 1160 m

4 0
3 years ago
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