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Phantasy [73]
3 years ago
10

A ball on the end of a rope is moving in a vertical circle near the surface of the earth. Point A is at the top of the circle; C

is at the bottom. Points B and D are exactly halfway between A and C. Which one of the following statements concerning the A boy is whirling a stone at the end of a string around his head. The string makes one complete revolution every second, and the tension in the string is FT. The boy increases the speed of the stone, keeping the radius of the circle unchanged, so that the string makes two complete revolutions per second. What happens to the tension in the sting?
Physics
1 answer:
mrs_skeptik [129]3 years ago
3 0

Answer:

Tension in the string will increase

Explanation:

As we know that tension in the string at any angle with the vertical is given as

T - mgcos\theta = m\omega^2 R

now we have

T = mgcos\theta + m\omega^2 R

also we know that

angular speed of the stone is directly depending on the time period of the motion

so it is given as

\omega = \frac{2\pi}{T}

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s

so the angular speed would be doubled

So here we can say that

tension in the string will increase when we will increase the frequency of revolution.

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A 400.0-m-wide river flows from west to east at 30.0 m/min. Your boat moves at 100.0 m/min relative to the water no matter which
Ratling [72]

Answer:

a. boat would drift towards a NE direction

b. 104.4m/min * t

c. false

d. point downstream of c, -79.19 deg NE

e. 3.898mins

f. 389.8m

Explanation:

a. downstream is a location with respect to the direction of the flow of the river, in this case from west to east. the boat would drift towards the NE direction due to the direction of water current.

b. lets say it takes time t to travel across the river, the resultant speed of boat would be given as Rs= sqrt ( 100/min^2 + 30m/mins^2) = 104.4m/mins

distance traveled would be 104.4 *t

c. false, the direction of boat wouldnt remain perpendicular the entire crossing but would drift toward the NE bearing provided the driver doesnt change or steer the boat.

d. a. taking the definition of downstream in question (a), your boat would land at a point x downstream to c

   b. to land at c boat must be inclined at an angle of -79.19 deg NE, given by using the width of river and the distance of b to c.

                            cosФ= 75/ 400

                                  Ф= 79.19°NE to get to c you must position boat at an angle of -79.19 deg from your initial bearing.

e. speed = distance/ time, time = distance / speed.

aiming toward point c, speed of boat is the resultant speed of boat with respect to the speed of water, using Pythagoras also

speed=  \sqrt{ 100^{2} + 30^{2} } = 104m/mins

distance to move from point a to c is given also given by

distance = \sqrt{400^{2} + 75^{2} }= 406.97m

time = 406.97/ 104.4 = 3.89mins...

f. on the ground the observer see the boat speed as 100m/min.

distance traveled is given by speed * time= 100 * 3.89= 389m

g. speed of boat observed would be 100m/min

5 0
3 years ago
Which of the following is likely the best electrical insulator
Vesnalui [34]
Is a vaccum cleaner there is no choices so here is one 
5 0
3 years ago
Arrange the objects in order from greatst to least of potential energy assume that gravity is constant
aleksley [76]

Answer:

Water > Box of books > Stone > Ball

Explanation:

We'll begin by calculating the potential energy of each object. This can be obtained as follow:

For stone:

Mass (m) = 15 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 3 m

Potential energy (PE) =?

PE = mgh

PE = 15 × 10 × 3

PE = 450 J

For water:

Mass (m) = 10 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 9 m

Potential energy (PE) =?

PE = mgh

PE = 10 × 10 × 9

PE = 900 J

For ball:

Mass (m) = 1 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 20 m

Potential energy (PE) =?

PE = mgh

PE = 1 × 10 × 20

PE = 200 J

For box of books:

Mass (m) = 25 Kg

Acceleration due to gravity (g) = 10 m/s²

Height (h) = 2 m

Potential energy (PE) =?

PE = mgh

PE = 25 × 10 × 2

PE = 500 J

Summary:

Object >>>>>>>> Potential energy

Stone >>>>>>>>> 450 J

Water >>>>>>>>> 900 J

Ball >>>>>>>>>>> 200 J

Box of books >>> 500 J

Arranging from greatest to least, we have:

Object >>>>>>>> Potential energy

Water >>>>>>>>> 900 J

Box of books >>> 500 J

Stone >>>>>>>>> 450 J

Ball >>>>>>>>>>> 200 J

Water > Box of books > Stone > Ball

4 0
3 years ago
What was done to protect the ozone layer?
svp [43]

Answer:

a. Ban Freon, aerosol cans, and refrigerants

Explanation:

5 0
2 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
3 years ago
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