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Vadim26 [7]
3 years ago
14

When you traveling downhill on a roller coaster, you experience a light feeling in your stomach. why

Physics
1 answer:
lesya [120]3 years ago
7 0
Because of the gravitational pull when we go up the gravity pulls u down
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Oo hi granger ru online here i have a doubt in physics . ​
NISA [10]

Answer:

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Explanation:

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3 years ago
No sistema representado abaixo, a massa do bloco A é igual a 2 kg, a do bloco B é igual a 8 kg.
blagie [28]

Answer:

Explanation:19,2 or 0/4 or 5 or 40,4

6 0
3 years ago
A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s
lisabon 2012 [21]

Answer:

Magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Given

Contact Time = t = 0.05 seconds

Mass (of ball) = 0.80kg

Initial Velocity = u = 25m/s

Final Velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is given by;

F = ma

Where m = 0.8kg

a = Average Acceleration

a = (u + v)/t

a = (25 + 25)/0.05

a = 50/0.05

a = 1000m/s²

Average Force = Mass * Average Acceleration

Average Force = 0.8kg * 1000m/s²

Average Force = 800kgm/s²

Average Force = 800N

Hence, the magnitude of the average force exerted on the wall by the ball is 800N

3 0
3 years ago
How is energy sent from Earth to space?
Nostrana [21]
It might be radiation and reflection but I’m not sure
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3 years ago
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8. An effort force of 15 Newtons is applied to an ideal pulley system to lift up a 16 Newton object. If the effort force is exer
Sonbull [250]

Answer:

the distance that the object is raised above its initial position is 5.625 m.​

Explanation:

Given;

applied effort, E = 15 N

load lifted by the ideal pulley system, L = 16 N

distance moved by the effort, d₁ = 6 m

let the distance moved by the object = d₂

For an ideal machine, the mechanical advantage is equal to the velocity ratio of the machine.

M.A = V.R

M.A = \frac{Load}{Effort} = \frac{L}{E} \\\\V.R = \frac{disatnce \ moved \  by \ the \ effort}{disatnce \ moved \  by \ the \ load} = \frac{d_1}{d_2} \\\\For \ ideal \ machine; \ M.A = V.R\\\\\frac{L}{E} = \frac{d_1}{d_2} \\\\d_2 = \frac{E \times d_1}{L} \\\\d_2 = \frac{15 \times 6}{16} \\\\d_2 = 5.625 \ m

Therefore, the distance that the object is raised above its initial position is 5.625 m.​

3 0
2 years ago
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