Three oxygen isotopes
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a) The work done is 920 J
b) The work done is 5200 J
Explanation:
a)
In this first part of the problem, the crate is moved horizontally at constant speed.
The work required in this case is given by
where
F is the magnitude of the force applied
d is the displacement of the crate
is the angle between the direction of the force and of the displacement
Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so
F = 230 N
The displacement is
d = 4.0 m
And the angle is , since the force is applied horizontally. Therefore, the work done is
b)
In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore
F = W = 1300 N
The displacement this time is again
d = 4.0 m
And the angle is , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is
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Answer:
The total time it is in the air for the ball is 1.6326 s
Given:
Initial velocity = 8
To find:
the total time it is in the air = ?
Formula used:
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
Solution:
A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.
The time taken by the ball to reach the maximum height is given by,
t =
Where t = time to reach maximum height
v = final velocity of the ball = 0 m/s
u = initial velocity of ball = 8 m/s
a = acceleration due to gravity = -9.8
Acceleration of gravity is taken as negative because ball is moving in opposite direction.
t =
t = 0.8163 s
Thus, time taken by the ball to reach the ground again = time taken to reach maximum height
So, Total time required for ball to reach ground = 2t = 2 × 0.8163
Total time required for ball to reach ground = 1.6326 s
The total time it is in the air for the ball is 1.6326 s
Answer:
Explanation:
For this exercise we must use the principle of conservation of energy
starting point. The proton very far from the nucleus
Em₀ = K = ½ m v²
final point. The point where the proton is stopped (v = 0)
Em_f = U = q V
where the potential is
V = k Ze / r²
Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching
Energy is conserved
Em₀ = Em_f
½ m v² = e ()
with this expression we can find the closest approach distance (r)