Answer:
C) The lander’s velocity decrease toward the reference.
Explanation:
took the test
Answer:
<em>The flux through the sphere will remain the same, and the magnitude of the electric field will increase by four times.</em>
Explanation:
The electric flux is the number of electric field, passing through a given area. It is proportional to the electric field strength and the area through which this field passes.
If the radius of the sphere is halved, the area of the sphere will reduce by square of the reduction, which will be four times. The electric field lines will become closer together, or technically increase by a fourth of its initial value. The resultant effect is that the electric flux will remain the same.
If originally,
Φ = EA cos∅
where Φ is the electric flux through the sphere
E is the electric field on the sphere
A is the area of the sphere.
If the area of the sphere is reduced to half, then,
the area reduces to A/4,
and the electric field increases to be 4E on the sphere.
The flux now becomes
Φ = 4E x A/4 cos∅
which reduces to
Φ = EA cos∅
which is the initial electric flux on the sphere.
1. The problem statement, all variables and given/known data A person jumps from the roof of a house 3.4 meters high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 meters. If the mass of his torso (excluding legs) is 41 kg. A. Find his velocity just before his feet strike the ground. B. Find the average force exerted on his torso by his legs during deceleration. 2. Relevant equations I can't even seem to figure that part out. Help please? 3. The attempt at a solution I don't know how to start this at all
Answer:
The magnitude of the force of friction equals the magnitude of my push
Explanation:
Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.
Let F = push and f = frictional force and f' = net force
F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0
So, F - f = 0
Thus, F = f
So, the magnitude of the force of friction equals the magnitude of my push.
Answer:
The magnitude of force is 
Solution:
As per the question:
The strength of Electric field due west at a certain point, 
Charge, Q = - 6 C
Now, the force acting on the charge Q in the electric field is given by:
Here, the negative sign indicates that the force acting is opposite in direction.
