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Airida [17]
3 years ago
7

A penny is placed on a rotating turntable. where on the turntable does the penny require the largest centripetal force to remain

in place?
Physics
1 answer:
Mama L [17]3 years ago
4 0

m = mass of the penny

r = distance of the penny from the center of the turntable or axis of rotation

w = angular speed of rotation of turntable

F = centripetal force experienced by the penny

centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

F = m r w²

in the above equation , mass of penny "m"  and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

hence greater the distance from center , greater will be the centripetal force to remain in place.  

So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.

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Answer:

-3.141593

Explanation:

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A police car is traveling north on a straight road at a constant 16.0 m/s. An SUV traveling north at 30.0 m/s passes the police
Nastasia [14]

Answer:

It will take 15.55s for the police car to pass the SUV

Explanation:

We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:

1. x=x_{0}+vt

2. x=x_{0}+v_{0}t+\frac{at^{2}}{2}

Since both cars will travel the same distance x, we can equal both formulas and solve for t:

vt = v_{0}t+\frac{at^2}{2}\\\\   16\frac{m}{s}t =30\frac{m}{s}t-\frac{1.8\frac{m}{s^{2}} t^{2}}{2}

We simplify the fraction present and rearrange for our formula so that it equals 0:

0.9\frac{m}{s^{2}} t^{2}-14\frac{m}{s}t=0 \\\\ t(0.9\frac{m}{s^{2}}t-14\frac{m}{s})=0

In the very last step we factored a common factor t. There is two possible solutions to the equation at t=0 and:

0.9\frac{m}{s^{2}}t-14\frac{m}{s}=0 \\\\  0.9\frac{m}{s^{2}}t =14\frac{m}{s} \\\\ t =\frac{14\frac{m}{s}}{0.9\frac{m}{s^{2}}}=15.56s

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at t=0 s (when the SUV passed the police car) and t=15.56s(when the police car catches up to the SUV)

8 0
3 years ago
3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of
Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

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2 years ago
A firecracker in a coconut blows the coconut into three pieces. Twopieces of equal mass fly off south and west, perpendicular to
loris [4]

Answer:

V = 13m/s

North-East (i.e. 45 degree from North)

Explanation:

This question deals with the idea of momentum.

Since the directions of a compass are fixed, we can take south as horizontal and west as vertical. Since both pieces of coconut are of equal mass, then the resultant of the two pieces would be in between South and West and the third piece would be opposite this direction and be in the North-East (i.e. 45 degree from North) direction

To find the speed of the third piece, we first find the speed on the two pieces of the coconut in terms of V_{x} (horizontal component of velocity) and V_{y} (vertical component of velocity)

We have

V_{y} = Velocity in South direction = 18m/s

m_{y} = Mass of piece in South direction = m

V_{x} = Velocity in West direction = 18m/s

m_{x} = Mass of piece in West direction = m

V_{ry} = Velocity of Third Piece in North direction = unknown

V_{rx} = Velocity of Third Piece in East direction = unknown

m = Mass of third piece = 2m

mV_{ry}  = m_{y} V_{y} \\ 2mV_{ry}  = m18\\ V_{ry} = 9

mV_{rx}  = m_{x} V_{x} \\ 2mV_{rx}  = m18\\ V_{rx} = 9

Since we now know the horizontal and vertical component of the velocity of the third piece of coconut we find the resultant velocity. Since we know the direction is North-East, we can imagine a right angled triangle with base of 9 and height 9 and the hypotenuse equal to the resultant velocity .

We can simply apply the Pythagoras theorem and find the hypotenuse

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