Answer:
E
= -4556.18 N/m
Explanation:
Given data
u = 3.6×10^6 m/sec
angle = 34°
distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in
Question)
from the projectile motion the horizontal distance traveled by electron is
x = u×cosA×t
⇒t = x/(u×cos A)
We also know that force in an electric field is given as
F = qE
q= charge , E= strength of electric field
By newton 2nd law of motion
ma = qE
⇒a = qE/m
Also, y = u×sinA×t - 0.5×a×t^2
⇒y = u×sinA×t - 0.5×(qE/m)×t^2
if y = 0 then
⇒t = 2mu×sinA/(qE) = x/(u×cosA)
Also, E = 2mu^2×sinA×cosA/(x×q)
Now plugging the values we get
E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})
E
= -4556.18 N/m
Answer:
The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg.
I would think 10 but I would have to see the picture
Explanation:
speeds = distance/time
=100/16
=6.25m/s
second speed is;
400/44 =9.09
av. speed = total speed /n
= (6.25+9.09)/2
=7.67 m/s
Given parameters:
Speed of the car = 50m/s
Time = 10s
Unknown:
Distance traveled before the car stopped = ?
We need to establish speed - time - distance relationship.
Speed is the rate of change of distance with time. It is a scalar quantity .
Mathematically;
Speed = 
Distance = speed x time
Input the parameters and solve;
Distance = 50 x 10 = 500m
The car traveled 500m before it stopped.
