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hammer [34]
3 years ago
5

6.) What force would evetually cause the ball to come to a stop after leaving the ramp​

Physics
1 answer:
solniwko [45]3 years ago
7 0

\colorbox{aqua} A\colorbox{aqua} N\colorbox{aqua} S\colorbox{aqua} W\colorbox{aqua} E\colorbox{aqua} R\colorbox{aqua} S

Forces of inertia

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Does Pascals law apply to solids?​
monitta
No it does not . That is the answer
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A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow
ale4655 [162]

Answer: The riders are subjected to 11.5 revolutions per minute

Explanation: Please see the attachments below

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3 years ago
Regions in the milky way where density waves have caused gas clouds to crash into each other are called
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7 0
3 years ago
Q.Solve the following circuit find total resistance RT. Also find value of voltage across resister RC.
Drupady [299]

Answer:

14.57 ohms

Explanation:

Here in the figure ,Rb & R₄are in series  & also  Rc & R₅ are in series. As they are in series , ( Rb + R₄ ) & (Rc & R₅) are in parallel . So the equivalent resistance in that branch = ( 2 + 18 ) ║ ( 3 + 12 )

                                          = 20 ║ 15

                                          = (20×15) / (20 + 15)

                                          = 8.57 ohms

Also Ra ( 6 ohm ) is in series with that branch ,. So the equivalent resistance of the whole circuit = 8.57 + 6 = 14.57 ohms.

5 0
3 years ago
Cole is riding a sled with initial speed of 5 m/s from west to east. the frictional force of 50 n exists due west. the mass of t
stepan [7]
We can calculate the acceleration of Cole due to friction using Newton's second law of motion:
F=ma
where F=-50 N is the frictional force (with a negative sign, since the force acts against the direction of motion) and m=100 kg is the mass of Cole and the sled. By rearranging the equation, we find
a= \frac{F}{m}= \frac{-50 N}{100 kg}=-0.5 m/s^2

Now we can use the following formula to calculate the distance covered by Cole and the sled before stopping:
a= \frac{v_f^2-v_i^2}{2d}
where
v_f=0 is the final speed of the sled
v_i=5 m/s is the initial speed
d is the distance covered

By rearranging the equation, we find d:
d= \frac{v_f^2-v_i^2}{2a}= \frac{-(5 m/s)^2}{2 \cdot (-0.5 m/s^2)}=25 m
3 0
3 years ago
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