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Bezzdna [24]
3 years ago
10

a man is 9 m behind a dorr of train when it starts moving with a=2ms^-1. how far the man have to run and after what time will he

catch the train. what is his full speed
Physics
1 answer:
Naya [18.7K]3 years ago
8 0
Is their a multiple choice to choose from I'm not sure the answer I got is even right.
That would be very helpful.
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The amount of time required for 2 successive wave crests to pass a fixed point is called wave ________.
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7 0
3 years ago
A student was traveling to see his grandmother who lives 15 miles north oh his home he started from rest and maintained a pace o
N76 [4]

Explanation:

Given parameters:

Distance  = 15miles north = 24140.2m

Initial velocity  = 0m/s

Final velocity  = 4m/s

Unknown:

Speed, velocity and acceleration = ?

Solution:

The speed is the distance divide by time. It is a scalar quantity and has no directional attribute.

  Speed  = \frac{distance }{time}  

  The speed of the student is 4m/s

Velocity is the displacement divided by time. It is a vector quantity which specifies the direction and magnitude;

    Velocity  = \frac{displacement}{time}  

   The velocity of the student is 4m/s due north

Acceleration is the change in velocity with time;

     To find the acceleration, we use

      v²  = u² + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

     4² = 0² + 2x a x 24140.2

       a  = \frac{16}{2 x 24140.2}  = 0.00033m/s²

3 0
3 years ago
The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
Vedmedyk [2.9K]

Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

Coefficient of static friction is, \mu = 0.214

Angle of inclination is, \theta = 31.5°

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

Therefore, the  tension in the rope is 229.37 N.

3 0
3 years ago
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