Answer:
The airliner travels 1.65 km along the runway before coming to a halt.
Explanation:
Given
Resistive forces = (2.90 × 10⁵) N = 290000 N
Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg
Velocity of airliner = 75 m/s
Let the distance over moved by the airliner be equal to d
According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.
Work done by the resistive forces = (290000) × d = (290,000d) J
Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J
290000d = 478,125,000
d = (478,125,000/290,000)
d = 1648.7 m = 1.65 km
Hope this helps!!!
Answer:
514 cal
Explanation:
In order to calculate the lost heat by the amount of water you first take into account the following formula:
(1)
Q: heat lost by the amount of water = ?
m: mass of the water
c: specific heat of water = 1cal/g°C
T2: final temperature of water = 11°C
T1: initial temperature = 12°C
The amount of water is calculated by using the information about the density of water (1g/ml):
Then, you replace the values of all parameters in the equation (1):
The amount of water losses a heat of 514 cal
Answer:
(a) 160000 kV/m
(b) 1336 keV
Explanation:
(a) magnetic filed, B = 10 T
energy of electron, E = 740 eV
mass of electron, m = 9.1 x 10^-31 kg
Let v be the velocity of electron.
E = 1/2 mv^2
740 x 1.6 x 10^-19 = 0.5 x 9.1 x 10^-31 x v^2
v = 1.6 x 10^7 m/s
v = E / B
E = v x B = 1.6 x 10^7 x 10 = 16 x 10^7 V/m
E = 160000 kV/m
(b) E = 16 x 10^7 V/m
B = 10 T
Let v be the velocity of protons.
v = E / B = 16 x 10^7 / 10 = 1.6 x 10^7 m/s
Kinetic energy of proton, E = 1/2 mv^2
= 0.5 x 1.67 x 10^-27 x 1.6 x 1.6 x 10^14
= 2.14 x 10^-13 J = 1336000 eV = 1336 keV
Answer:
b) Momentum decreased
Explanation:
The relative velocity ( with repect to hand) of ball is decreased by moving hand with some velocity in the direction of ball's velocity.
