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Refer to the diagram shown below.
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Assume that air resistance is ignored.
Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²
The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²
Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s
The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m
Answer: 319.4 m (nearest tenth)
Explanation:
It is given that,
Magnetic field, B = 0.1 T
Acceleration,
Charge on electron,
Mass of electron,
(a) The force acting on the electron when it is accelerated is, F = ma
The force acting on the electron when it is in magnetic field,
Here,
So,
Where
v is the velocity of the electron
B is the magnetic field
v = 341250 m/s
or
So, the speed of the electron is
(b) In 1 ns, the speed of the electron remains the same as the force is perpendicular to the cross product of velocity and the magnetic field.