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Setler [38]
3 years ago
13

Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl

ect the weight of a and
b. the coefficient of static friction for all contacting surfaces is μs=0.35. neglect friction at the rollers.
Physics
1 answer:
BARSIC [14]3 years ago
6 0
<span>b. The coefficient of static friction for all contacting surfaces is μs=0.35. neglect friction at the rollers.

</span>
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A 55.0-kg skydiver drew falls for a period of time before opening his parachute. what is his kinetic energy when he reaches a ve
Jlenok [28]
Mass (m)=55kg

acceleration (a)=9.81 m/s^2, this is the acceleration due to gravity.

initial velocity=0m/s. The skydiver doesn’t start with any speed because she is on the plane or helicopter.

final velocity=16m/s This is the velocity (speed) the skydiver reaches

The equation we use is KE=.5mv^2
Kinetic energy=.5 mass x velocity^2

KE=.5(55kg)(16m/s)^2
KE=.5(55kg)(256m/s)
KE=.5(14080J)

J=Joules

KE=7040J

Kinetic energy is 7040 Joules (J)

Hope this helps
3 0
4 years ago
If you place a paper clip very close to a magnet, the paper clip
Alik [6]
The answer is “D. all of the above”!

Metal from the paper clip is attracted to the magnet, so it will naturally move toward and stick to the magnet. This will cause the paper clip to temporarily become a magnet for other metals. I hope this helped!
7 0
2 years ago
The charge of an electron is
AlladinOne [14]
Proton positive; electron negative; neutron no charge<span>. </span>The charge<span> on the proton and </span>electron<span> are exactly the same size but opposite. The same number of protons and </span>electrons<span> exactly cancel one another in a neutral atom.
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6 0
3 years ago
The brakes on a 15,680 N car exert a stopping force of 640 N.
Murrr4er [49]

Answer:

1568 Kg is the answer

Explanation:

6 0
3 years ago
14. A rocket is shot up into the air and then comes back down and hits the ground 9.2 second later.
sineoko [7]

Answer:

105.8 m

46 m/s

Explanation:

From the time the rocket is launched to the time it reaches its maximum height:

v = 0 m/s

a = -10 m/s²

t = 9.2 s / 2 = 4.6 s

Find: Δy and v₀

Δy = vt − ½ at²

Δy = (0 m/s) (4.6 s) − ½ (-10 m/s²) (4.6 s)²

Δy = 105.8 m

v = at + v₀

0 m/s = (-10 m/s²) (4.6 s) + v₀

v₀ = 46 m/s

3 0
4 years ago
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