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aev [14]
3 years ago
9

8TH GRADE SCIENCE I NEED NOW DO NOT SKIP

Physics
1 answer:
yan [13]3 years ago
7 0

Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,

So it will be= 50(9.8-2)

=50(7.8)= 390N

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Pretend you (80 kg) are making repairs on the outside of the International Space Station. You are floating 20 meters away from t
djyliett [7]

Answer:

32 seconds

Explanation:

m1 = 80 kg

m2 = 10 kg

v2 = 5m/s

According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:

m_{1} v_{1} =m_{2} v_{2} \\80v_{1} =10*5 \\v_{1} = 0.625\ m/s

Since the space station is 20 meters away, the time taken to reach it is given by:

t = \frac{20}{0.625}\\t=32\ s

It takes you 32 seconds to reach the station.

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3 years ago
How do Earth's and Jupiter's orbits compare?
shutvik [7]
Earth is smaller and have 1 moon so it rotates faster than jupiter and it have 6 moons
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3 years ago
Which of Newton’s Laws involves mass and acceleration? Question 1 options: 3rd 1st 2nd All of them
Feliz [49]
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6 0
3 years ago
Read 2 more answers
What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your
torisob [31]
<span>Radius distance from origin to particle = √ (2²+1²) = √5 m = R 
I = MR² = (0.200)(5) = 1.00 kg-m² 
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V = 3.0 m/s 
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s 
w = [V component / R] = 1.3433/√5 = 0.601 rad/s 
size of angular momentum of particle relative to origin = Iw = (1.00)(0.601) = 0.601 kgm²/s</span><span>


i hope I'm right</span>
6 0
3 years ago
A baseball bat is 32 inches (81.3 cm) long and has a mass of 0.96 kg. Its center of mass is 22 inches (55.9 cm) from the handle
s344n2d4d5 [400]

Answer:

0.24 kgm²

Explanation:

L = length of the bat = 81.3 cm = 0.813 m

m  = mass of the bat = 0.96 kg

d  = distance of the center of mass of bat from the axis of rotation = 55.9 cm = 0.559 m

T  = Period of oscillation = 1.35 sec

I = moment of inertia of the bat

Period of oscillation is given as

T = 2\pi \sqrt{\frac{I}{mgd}}

1.35 = 2(3.14) \sqrt{\frac{I}{(0.96)(9.8)(0.559)}}

I = 0.24 kgm²

6 0
3 years ago
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