(3) 8.3 N/kg. The gravitational field strength at a point is the force per unit mass exerted on a mass placed at that point. So at the point where the Hubble telescope is, it is (9.1 x 10^4)N/(1.1 x 10^4 kg) = 8.3 N/kg
Fam
Answer:
b) q large and m small
Explanation:
q is large and m is small
We'll express it as :
q > m
As we know the formula:
F = Eq
And we also know that :
F = Bqv
F = 
Bqv = 
or Eq = 
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.
Answer:
zero
Explanation:
Acceleration is a measure of the rate of change of velocity. If the velocity is unchanging, its rate of change is zero.
The acceleration is zero.