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Phantasy [73]
3 years ago
11

Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of

50 kilograms. What can Paul do to increase the bike’s kinetic energy?
A. He can let Ivan off at the next stop.
B. He can pedal harder to increase the rate to 10 meters/second.
C. He can reduce the speed to 3 meters/second.
D. He can pick up a third rider.
Physics
2 answers:
Tresset [83]3 years ago
8 0

Answer: Well they could go down a hill to gain more kinetic energy, or the answer can just be B. He can pedal harder to increase the rate to 10 meters/second. I hope I helped you.

GaryK [48]3 years ago
4 0

Answer:

the answer is probably reducing the speed, hope i helped! :D

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The frequency of a wave becomes higher due to the object moving at a fast pace coming towards you with shorter wavelengths  (depending on the speed) aka the Doppler Effect. 
Hope this helps
3 0
3 years ago
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The escape speed from a very small asteroid is only 24 m/s. If you throw a rock away from the asteroid at a speed of 30 m/s, wha
svp [43]

Given Information:  

Initial speed of rock = vi = 30 m/s  

escape speed of the asteroid = ve = 24 m/s  

Required Information:  

final speed of rock = vf = ?

Answer:  

vf = 18 m/s

Explanation:  

As we know from the conservation of energy

KEf + Uf = KEi + Ui

Where KE is the kinetic energy and U is the potential energy

0 + 0 = ½mve² - GMm/R

When escape speed is used, KEf is zero due to vf being zero. Uf is zero because the object is very far away from mass M, therefore, the equation becomes

GMm/R = ½mve²

m cancels out

GM/R = ½ve²

GM/R = ½(24)²

GM/R = 288

KEf + Uf = KEi + Ui

½mvi² + 0 =  ½vf² - GMm/R

m cancels out

½vi² =  ½vf² - GM/R

Substitute the values

½(30)² =  ½vf² - (288)

½vf² = 450 - 288

vf² = 2(162)

vf = √324

vf = 18 m/s

Therefore, the final speed of the rock is 18 m/s

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3 years ago
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Compare the properties of sodium chloride and sand
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3 years ago
What is the name of the theory describing how the lithosphere is broken into segments, or plates, which "float" on the asthenosp
kogti [31]

Answer:

C. Plate Tectonics

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4 0
3 years ago
Question 1 Gary is on the space shuttle. It takes off and lifts him to a height of 300 km above Earth's surface. a. How has Gary
balandron [24]
A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.

b) The weight of an object is given by
W=mg
where
m is the mass
g= \frac{GM}{r^2} is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.

At the Earth's surface, g=9.81 m/s^2, so Gary's weight is
W=mg=9.81 m  (1)
where m is Gary's mass.

Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is 
R=6370 km
So the distance of Gary from the Earth's center is
r=R+h=6370 km+300 km =6670 km = 6.67 \cdot 10^6 m

The Earth's mass is M=5.97 \cdot 10^{24} kg, so the gravitational acceleration is
g'=G \frac{M}{r^2}= (6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2} )\frac{5.97 \cdot 10^{24} kg}{(6.67 \cdot 10^6 m)^2}=8.95 m/s^2

Therefore, Gary's weight at 300 km above Earth's surface is 
W' = mg' = 8.95 m (2)

If we compare (1) and (2), we find that Gary's weight has changed by
\frac{W'}{W}= \frac{8.95 m}{9.81 m}=0.91
So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.
6 0
3 years ago
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