Only the component of force perpendicular to the direction motion does work. true or false
1 answer:
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The strength of the gravitational field is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
.
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
Answer:
B=9.1397*10^-4 Tesla
Explanation:
To find the velocity first we put kinetic energy og electron is equal to potential energy of electron
K.E=P.E
where :
m is the mass of electron
v is the velocity
V is the potential difference
eq 1
Radius of electron moving in magnetic field is given by:
eq 2
where:
m is the mass of electron
v is the velocity
q=e=charge of electron
B is the magnitude of magnetic field
Put v from eq 1 into eq 2
B=9.1397*10^-4 Tesla