Answer:
The magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
Explanation:
The initial centripetal acceleration, a of the race-car around the circular track of radius , R with a linear speed v is a = v²/R.
When the linear speed of the race-car increases to v' = 4v, the centripetal acceleration a' becomes a' = v'²/R = (4v)²/R = 16v²/R.
So the centripetal acceleration, a' = 16v²/R.
To know how much the magnitude of the car's centripetal acceleration changes, we take the ratio a'/a = 16v²/R ÷ v²/R = 16
a'/a = 16
a' = 16a.
So the magnitude of the centripetal acceleration increases by 16 times when the linear speed increases by 4 times.
It Irregularly Reverses. I got the answer right on the test.
Answer: acceleration = slope graph velocity vs time
Explanation: if you have the graph of velocity vs time , the slope of that graph equals the acceleration of our object assuming constant acceleration...but remenber por a real object is really hard to keep constant acceleration
Answer:
We can use 2 g H = v2^2 - v1^2 or
v2^2 = 2 g H + v1^2
Since 88 ft/sec = 60mph we have 30 mph = 44 ft/sec
The object will return with the same speed that it had initially so the object
starts out with a downward speed of 44 ft/sec
Then v2^2 = 2 * 32 ft/sec^2 * 160 ft + 44 (ft/sec)^2
v2^2 = (2 * 32 * 160 + 44^2) ft^2 / sec^2 = 12180 ft^2/sec^2
v2 = 110 ft/sec
Answer:
Explanation:
The charge on one object, 
The distance between the charges, r = 0.22 m
The force between the charges, F = 4,550 N
Let q₂ is the charge on the other sphere. The electrostatic force between two charges is given by the formula as follows :
So, the charge on the other sphere is 
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