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bagirrra123 [75]
3 years ago
9

An object of mass 3.00 kg, moving with an initial velocity of 5.05 m/s, collides with and sticks to an object of mass 2.76 kg wi

th an initial velocity of -3.66 m/s. Find the final velocity of the composite object.
Physics
1 answer:
Lynna [10]3 years ago
6 0

Answer:

0.752 m/s

Explanation:

m1 = 3.00kg

u1 = 5.05m/s

m2 = 2.76kg

u2 = -3.66m/s

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

3(5.05) + 2.76(-3.66) = (5.05+2.76)v

15.15 - 9.2736 = 7.81v

5.8764 = 7.81v

v = 5.8764/7.81

v = 0.752m/s

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A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm
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Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

t=5\cdot 10^{-5}s

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

c=3.0\cdot 10^8 m/s

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

d=vt (1)

where

v=c=3.0\cdot 10^8 m/s

t=5\cdot 10^{-5}s

d=2L, where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m

7 0
4 years ago
A child rides the train at Disneyland which takes him in a loop around the amusement park at a speed of 3 m/s. Which statement b
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1. What is work done in holding a 15kg suitcase while waiting for a bus for 15 minutes?
Anettt [7]
The man is holding the suitcase at the same height above the surface of earth. So the gravitation potential energy remains the same. 

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3 years ago
A paintball is shot horizontally in the positive x direction at time t after the ball is shot it is 4 cm to the right and 4 cm b
Artist 52 [7]

<u>Answer:</u>

At time 2t the paint ball is at 8 cm to the right and 16 cm to the bottom

<u>Explanation:</u>

 We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Considering the horizontal motion of paint ball

    Distance traveled during time t = 4 cm

    Initial velocity = u m/s

   Acceleration = 0 m/s^2

So 4 = u*t+\frac{1}{2} *0*t^2\\ \\ u = \frac{4}{t}

Now at time 2t,

  S= u*2t+\frac{1}{2} *0*(2t)^2\\ \\=\frac{4}{t} *2t\\ \\ =8cm

  So horizontal distance traveled in time 2t = 8 cm to the right

Now considering the vertical motion of paint ball

  Distance traveled during time t = 4 cm

    Initial velocity = 0 m/s

   Acceleration = -g m/s^2

4=0*t-\frac{1}{2} *g*t^2\\ \\ t^2=\frac{-8}{g}

At time 2t,

     S=0*2t-\frac{1}{2} *g*(2t)^2\\ \\ =-\frac{1}{2} *g*4*\frac{-8}{g}\\ \\ =16 cm

 So vertical distance traveled in time 2t = 16 cm to the bottom

 

6 0
3 years ago
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