Answer:
Time to pass the train=0.05 h
How far the car traveled in this time=4.75 Km
Explanation:
We have that the train and the car are moving in the same direction, the difference between the speed of the vehicles is:
We will use this difference in the speed of the car an train to calculate how much time take the car to pass the train. For this we have that the train is 1km long and the car is moving with a speed of 20km/h (we use this value because is the speed that the car have in advantage of the train) then for a movement with a constant speed we have:
Where x is the distance, t is the time and v is the speed. using the data that we have:
This is the time that the car take to pass the train. Now to calculate how far the car have traveled in this time we have to considered the speed of 95Km/h of the car, then:
Answer:
the potential energy of this body is 245 J.
Explanation:
Given;
mass of the body, m = 250 g = 0.25 kg
height from which the body was dropped, h = 100 m
acceleration due to gravity, g = 9.8 m/s²
The potential energy of this body is calculated as;
P.E = mgh
substitute the given values and solve for the potential energy of this body;
P.E = 0.25 x 9.8 x 100
P.E = 245 J.
Therefore, the potential energy of this body is 245 J.
Answer:
1.25 m/s
Explanation:
Given,
Mass of first ball=0.3 kg
Its speed before collision=2.5 m/s
Its speed after collision=2 m/s
Mass of second ball=0.6 kg
Momentum of 1st ball=mass of the ball*velocity
=0.3kg*2.5m/s
=0.75 kg m/s
Momentum of 2nd ball=mass of the ball*velocity
=0.6 kg*velocity of 2nd ball
Since the first ball undergoes head on collision with the second ball,
momentum of first ball=momentum of second ball
0.75 kg m/s=0.6 kg*velocity of 2nd ball
Velocity of 2nd ball=0.75 kg m/s ÷ 0.6 kg
=1.25 m/s
Answer:
Explanation:
Charge on uranium ion = charge of a single electron
= 1.6 x 10⁻¹⁹ C
charge on doubly ionised iron atom = charge of 2 electron
= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C
Let the required distance from uranium ion be d .
force on electron at distance d from uranium ion
= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²
force on electron at distance 61.10 x 10⁻⁹ - r from iron ion
= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²
For equilibrium ,
9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²
2 d² = (61.10 x 10⁻⁹ - r )²
1.414 r = 61.10 x 10⁻⁹ - r
2.414 r = 61.10 x 10⁻⁹
r = 25.31 nm .
