Answer:
The acceleration due to gravity is
times the value of g at the Earth’s surface.
(D) is correct option.
Explanation:
Given that,
Radius = 4000 miles
We need to calculate the gravitational force at surface
Gravitational force on the mass m on the surface of the earth
At r = R

....(I)
We need to calculate the gravitational force at height
Gravitational force on a mass m from the center of the earth,
At r = R + R = 2 R

....(II)
Dividing equation (II) by equation (I)


Hence, The acceleration due to gravity is
times the value of g at the Earth’s surface.
Explanation:
It is given that, a long, straight wire is surrounded by a hollow metal cylinder whose axis coincides with that of the wire.
The charge per unit length of the wire is
and the net charge per unit length is
.
We know that there exist zero electric field inside the metal cylinder.
(a) Using Gauss's law to find the charge per unit length on the inner and outer surfaces of the cylinder. Let
are the charge per unit length on the inner and outer surfaces of the cylinder.
For inner surface,



For outer surface,



(b) Let E is the electric field outside the cylinder, a distance r from the axis. It is given by :


Hence, this is the required solution.
<span>B: adds aesthetic value to the landscape. Think about it, out of all your options, that's the one that doesn't really help anything.
And I took the test, so take my word for it.</span>
Answer:
23 m/s downward
__________________________________________________________
<em>Taking the downward direction as positive</em>
<u>We are given:</u>
Initial velocity of the marble (u) = 0 m/s
Time interval (t) = 2.3 seconds
Final velocity (v) = x m/s
<u>Solving for the Final velocity:</u>
<u>Acceleration of the Marble:</u>
We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s
<u>Final velocity:</u>
v = u + at [First equation of motion]
x = 0 + (10)(2.3) [replacing the given values]
x = 23 m/s
Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction
m = mass = 5 kg
= initial velocity = 100 m/s
= final velocity = ?
I = impulse = 30 Ns
Using the impulse-change in momentum equation
I = m(
-
)
30 = 5 (
- 100)
= 106 m/s