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Irina-Kira [14]
3 years ago
11

You compress a spring by a certain distance. then you decide to compress it further so that its elastic potential energy increas

es by another 50%. what was the percent increase in the spring's compression distance?
Physics
1 answer:
Likurg_2 [28]3 years ago
5 0
Elastic potential energy = 1/2 k * change of x^2
k- coefficient
x - change in length.
to increase energy 1.5 times you have to change x (compress) into \sqrt{1.5}  times (it's abot 1.22 or 22%)
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There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

Orbital radius r = 4.28  × 10⁵km

Orbital period  T = 28.9days

T = 28.9  × 24 × 60 × 60

   = 2,496,960s

Angular momentum of the moon about the planet

L = mvr

L = mr²w

L = mr^2\frac{2\pi }{T} \\\\L = \frac{9.5 \times 10^2^2 \times(4.28\times10^8)^2\times2\times3.14}{2496960} \\\\L = 4.389.5 \times 10^3^4kgm^2/s

7 0
3 years ago
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Compare and contrast what happens to an object's motion when balanced or
mario62 [17]

Answer: The motions shifts.

5 0
3 years ago
A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0
3 years ago
A ball is thrown at 60 degrees and lands in 18.5 seconds what is the velocity of the ball at the start
n200080 [17]

Answer:

the initial velocity of the ball is 104.67 m/s.

Explanation:

Given;

angle of projection, θ = 60⁰

time of flight, T = 18.5 s

let the initial velocity of the ball, = u

The time of flight is given as;

T = \frac{2u\times sin(\theta)}{g} \\\\2u\times sin(\theta) = Tg\\\\u = \frac{Tg}{2\times sin(\theta)} \\\\u = \frac{18.5 \times 9.8}{2\times sin(60^0)} \\\\u = 104.67 \ m/s

Therefore, the initial velocity of the ball is 104.67 m/s.

3 0
3 years ago
PLEASE I NEED HELP ASAP!!!!!!!!!!
max2010maxim [7]

tooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo

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3 years ago
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