Answer:
63.750KeV
Explanation:
We are given that
Initial velocity of second electron,
Radius,
1 m=100 cm
Magnetic field,B=0.0370 T
We have to determine the energy of the incident electron.
Mass of electron,
Charge on an electron,
Velocity,
Using the formula
Speed of electron,
Speed of second electron,
Kinetic energy of incident electron=
Kinetic energy of incident electron=
Kinetic energy of incident electron=
1KeV=1000eV
Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction 
Centripetal force is given 
And frictional force is equal to 
For body to be move these two forces must be equal
So 
The angular acceleration of the blade when it's switched off is (-6800 rev/min) divided by (2.8 sec) = -2,428.6 rev/(min-sec) = -40.5 rev/sec^2 .
Answer:
3.1 m/s
Explanation:
First, find the time it takes for the cat to land. Take down to be positive.
Given:
Δy = 0.61 m
v₀ = 0 m/s
a = 9.81 m/s²
Find: t
Δy = v₀ t + ½ at²
(0.61 m) = (0 m/s) t + ½ (9.81 m/s²) t²
t = 0.353 s
Now find the horizontal velocity needed to travel 1.1 m in that time.
Given:
Δx = 1.1 m
a = 0 m/s²
t = 0.353 s
Find: v₀
Δx = v₀ t + ½ at²
(1.1 m) = v₀ (0.353 s) + ½ (0 m/s²) (0.353 s)²
v₀ = 3.1 m/s
