Answer:
V = 0.714m/s
Explanation:
Full solution calculation can be found in the attachment below.
From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.
Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.
See the attachment below for the solution calculation.
Answer:
He traveled 9km
Explanation:
To do this problem you need to use the equation which is Speed= distance/time and this problem gives you the speed which is 18 km/h and it gives you the time 1/2 hour so you write the equation 18= d/ 1/2 which his distance is 9km
Answer:
<em>2 m/s</em>
<em></em>
Explanation:
The electromagnetic flow-metre work on the principle of electromagnetic induction. The induced voltage is given as
where 
is the induced voltage = 2.88 mV = 2.88 x 10^-3 V
is the distance between the electrodes in this field which is equivalent to the diameter of the tube = 1.2 cm = 1.2 x 10^-2 m
is the velocity of the fluid through the field = ?
is the magnetic field = 0.120 T
substituting, we have
2.88 x 10^-3 = 0.120 x 1.2 x 10^-2 x
2.88 x 10^-3 = 1.44 x 10^-3 x
= 2.88/1.44 = <em>2 m/s</em>
Answer:
The ball stops instantaneously at the topmost point of the motion.
Explanation:
Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.
The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.
The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.
Answer:
a)
The direction will be negative direction.
b)
The direction will be positive direction.
Explanation:
Given that
q1 = +7.7 µC is at x1 = +3.1 cm
q2 = -19 µC is at x2 = +8.9 cm
We know that electric filed due to a charge given as
Now by putting the va;ues
a)
The net electric field
The direction will be negative direction.
As we know that electric filed line emerge from positive charge and concentrated at negative charge.
b)
Now
distance for charge 1 will become =5.5 - 3.1 = 2.4 cm
distance for charge 2 will become =8.9-5.5 = 3.4 cm
The net electric field
The direction will be positive direction.
