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DENIUS [597]
3 years ago
10

If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circular path were reduced,

what would happen?
Physics
2 answers:
Tanya [424]3 years ago
8 0

For PF STUDENTS:  The car would begin to move in the direction it was headed in a straight line.

saw5 [17]3 years ago
3 0
As the question states the reduction of the centripetal force is a reduction of the frictional force. By reducing these forces (they are the same force) the tires might not follow the circular path, this is the car could slip because the force would be insufficient to keep the circular path.
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I believe ROCKFALLS is the answer.
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3 years ago
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A tuning fork with a frequency of 335 Hz and a tuning fork of unknown frequency produce beats with a frequency of 5.3 when struc
Ganezh [65]

Answer:

Explanation:

Unknown fork frequency is either

335 + 5.3 = 340.3 Hz

or

335 - 5.3 = 329.7 Hz

After we modify the known fork, the unknown fork frequency equation becomes either

(335 - x) + 8 = 340.3

(335 - x)  = 332.3

x = 2.7 Hz

or

(335 - x) + 8 = 329.7

(335 - x) = 321.7

x = 13.3 Hz

IF the unknown fork frequency was 340.3 Hz,

THEN the 335 Hz fork was detuned to 335 - 2.7 = 332.3 Hz

IF the unknown fork frequency was 329.7 Hz,

THEN the 335 Hz fork was detuned to 335 - 13.3 = 321.7 Hz

3 0
3 years ago
A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha
dem82 [27]

Answer:

no, 3 porces is more tha 2 so the power between the 3 should be more than 2

Explanation:

4 0
3 years ago
Can someone please Help me with this? It’s Due today
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6 0
3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
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