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Bas_tet [7]
2 years ago
8

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

h a velocity of 103 m/s west. What is the final velocity of the block? Assume the block rests on a perfectly frictionless horizontal surface.
Physics
1 answer:
makvit [3.9K]2 years ago
7 0

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

            2.87 = -1.442 + 1.8 u

               u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

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You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0
2 years ago
Which of the following statements characterizing types of waves are true?
Tasya [4]

Answer:

a and b.

Explanation:

In general  types of wave

1. Transverse wave  :

    In these waves particle are vibrate perpendicular to motion of waves.

 Ex : Electromagnetic wave , Radio wave .

2. Longitudinal wave :

   In these waves particle are vibrate along the motion of waves.

 Ex : Sound wave

Mechanical wave  :

1 .These are transverse wave or Longitudinal wave or combination of them .

2.These waves required medium for propagation.

3. The particle are vibrate perpendicular to motion of waves.

So the option a and b are correct.

7 0
2 years ago
8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
Vlad1618 [11]

Answer:

B_o = 1.013μT

Explanation:

To find B_o you take into account the formula for the emf:

\epsilon=-\frac{d\Phi_b}{dt}=-\frac{dBAcos\theta}{dt}=-Acos\theta\frac{dB}{dt}

where you used that A (area of the loop) is constant, an also the angle between the direction of B and the normal to A.

By applying the derivative you obtain:

\epsilon=-Acos\theta (2\pi f) B_ocos(2\pi f t+ \alpha)

when the emf is maximum the angle between B and the normal to A is zero, that is, cosθ = 1 or -1. Furthermore the cos function is 1 or -1. Hence:

\epsilon=2\pi fAB_o=2\pi (100*10^3Hz)(\pi (0.1m)^2)B_o=19739.20Hzm^2B_o\\\\B_o=\frac{20*10^{-3}V}{19739.20Hzm^2}=1.013*10^{-6}T=1.013\mu T

hence, B_o = 1.013μT

6 0
2 years ago
A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

\epsilon = -\frac{\Delta \Phi_B}{\Delta t} (1)

where

\Delta \Phi_B is the variation of magnetic flux through the coil

\Delta t = 2.0 ms = 0.002 s is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

\Phi_B = BA

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2

At the beginning, the magnetic field is

B_i = 1.5 T

so the flux is

\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

\Phi_f = 0

so the variation of magnetic flux is

\Delta \Phi = 0-0.068 Wb=-0.068 Wb

And substituting into (1) we find the average emf in the coil

\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0
2 years ago
In an experiment to measure the wavelength of light using a double slit, it is found that the fringes are too close together to
Mandarinka [93]

Answer:

halve the slit separation

Explanation:

As we know that

In YDS experiment, the equation of fringe width is as follows

\beta = \frac{\lambda D}{d}

where,

D denotes the separation in the middle of screen and slits

d denotes the distance in the middle of two slits

And to increase the Δx we have to decrease the d i.e, the distance between the two slits

Hence, the first option is correct

4 0
2 years ago
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