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Thepotemich [5.8K]
2 years ago
7

Suppose you are on a cart that is moving at a constant speed v toward the left on a frictionless track. If you throw a massive b

all straight up (from your perspective), how will the speed of the cart change?
a. The speed of the cart will increase
b. The speed of the cart will decrease
c. The speed of the cart will not change
d. You need to know how fast the ball was thrown
Physics
1 answer:
Alex2 years ago
6 0
The correct answer is A
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There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think
dangina [55]

Answer:

4.44 rpm

Explanation:

\omega = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = \frac{3138000}{2}\ m

R = Radius of arm = 6 m

The acceleration due to gravity is given by

g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

a_c=\omega^2R

a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s

Converting to rpm

1\ rad/s=\frac{60}{2\pi}\ rpm

0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm

The angular speed of the arm is 4.44 rpm

8 0
3 years ago
a force of 50 newtons pulls a rope attached to a 150 newton sled across a horizontal surface at a constant velocity of 5 meters
lakkis [162]

Answer:

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3 0
3 years ago
An Olympic high jumper, with a mass of 82 kg, has a
Digiron [165]

Answer:

I don't really know

Explanation:

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8 0
2 years ago
11.
Hoochie [10]
C. is correct : ) it is applied when you push and that causes the desk to move
7 0
3 years ago
Read 2 more answers
This time particle A starts from rest and accelerates to the right at 65.5 cm/s
FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(65.5)t^2

d_1 = 32.75 t^2 cm

Now we know that B moves with constant speed so in the same time B will move to another distance

d_2 = 44 \times t

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

d_1 = 349 + d_2

32.75 t^2 = 349 + 44 t

on solving above kinematics equation we have

t = 4 s

4 0
3 years ago
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