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Dmitriy789 [7]
3 years ago
12

A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. the balls under

go a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle umax = 50. what was v0 ?
Physics
1 answer:
Novay_Z [31]3 years ago
7 0

Sorry but i dont know i was doing this for some point sorry

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Ok plz answer and tell me how to do it
kirza4 [7]
Answer: 25N

method: total force in the right hand direction is 100N and total force in the left hand direction is 125N. To get the net force, we add forces if they are in the same direction and substract if they are in opposite directions. since 100N and 125N are in opposite directions, we substract the larger value from the smaller value. Then we get 25N in the left hand direction as the final answer.
4 0
3 years ago
How much force is needed to slow down a 15 kg car traveling at 60 m/s to 15 m/s in 10 seconds?​
goldfiish [28.3K]

Answer:

The force needed to slow down the car is, F = 67.5 N

Explanation:

Given data,

The mass of the car, m = 15 kg

The initial velocity of the car, V = 60 m/s

The final velocity of the car, v = 15 m/s

The time period of deceleration, t = 10 s

The difference in the momentum of the car is,

                                     mV - mv = 15(60 - 15)

                                                    = 675 kg m/s

The rate of change in momentum of the car gives the force acting on it.

                                    F = (mV - mu) / t

Substituting the values,

                                   F = 675 / 10

                                      = 67.5 N

Hence, the force needed to slow down the car is, F = 67.5 N

3 0
3 years ago
A conductor shaped as a circular loop with a radius of 4.0 m is located in a uniform but changing magnetic field. If the maximum
bekas [8.4K]

Answer:

\frac{\delta B}{\delta t}= 0.0995 \  T/s

Explanation:

Given that :

The radius of the circular loop = 4.0 m

Maximum Emf E_{max} = 5.0 V

The  maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies can be determined via the expression;

E_{max} = Area (A) * \frac{\delta B}{\delta t}

E_{max} = \pi r^2 * \frac{\delta B}{\delta t}

5.0 = \pi * (4.0)^2 * \frac{\delta B}{\delta t}

5.0 = 50.27 * \frac{\delta B}{\delta t}

\frac{\delta B}{\delta t}= \frac{5}{50.27}

\frac{\delta B}{\delta t}= 0.0995 \  T/s

5 0
3 years ago
(b) If the object is at 330 feet and its instantaneous velocity is 3 feet per minute at 30 minutes, what is the approximate posi
ELEN [110]

Answer:

The final position is 36 feet.

Explanation:

initial position, d = 330 feet

speed, v = 3 feet per minute

time, t = 30 minute

now the time is 32 minute

time interval = 2 minute

So, the distance in 2 minutes is

d' = 2 x 3 = 6 feet

So, the final position is

D = 30 + 6 = 36 feet

8 0
3 years ago
Suppose that a rectangular toroid has 1,500 windings and a self-inductance of 0.060 H. If the height of the rectangular toroid i
V125BC [204]

Answer:

Current in the toroid will be I=17.32\times 10^{-3}A

Explanation:

We have given number of winding in rectangular toroid N = 1500

Self inductance of toroid L = 0.06 H

Magnetic energy stored in toroid E=9\times 10^{-16}J

We have to find the current in the toroid

Magnetic energy stored is equal to E=\frac{1}{2}Li^2

9\times 10^{-6}=\frac{1}{2}\times 0.06\times I^2

I=17.32\times 10^{-3}A

So current in the toroid will be I=17.32\times 10^{-3}A

5 0
4 years ago
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