-- The car starts from rest, and goes 8 m/s faster every second.
-- After 30 seconds, it's going (30 x 8) = 240 m/s.
-- Its average speed during that 30 sec is (1/2) (0 + 240) = 120 m/s
-- Distance covered in 30 sec at an average speed of 120 m/s
= <span> 3,600 meters .</span>
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The formula that has all of this in it is the formula for
distance covered when accelerating from rest:
Distance = (1/2) · (acceleration) · (time)²
= (1/2) · (8 m/s²) · (30 sec)²
= (4 m/s²) · (900 sec²)
= 3600 meters.
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When you translate these numbers into units for which
we have an intuitive feeling, you find that this problem is
quite bogus, but entertaining nonetheless.
When the light turns green, Andy mashes the pedal to the metal
and covers almost 2.25 miles in 30 seconds.
How does he do that ?
By accelerating at 8 m/s². That's about 0.82 G !
He does zero to 60 mph in 3.4 seconds, and at the end
of the 30 seconds, he's moving at 534 mph !
He doesn't need to worry about getting a speeding ticket.
Police cars and helicopters can't go that fast, and his local
police department doesn't have a jet fighter plane to chase
cars with.
Answer:
-54.12 V
Explanation:
The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.

The change in electrostatic potential energy
, of one point charge q is defined as the product of the charge and the potential difference.

Answer:
x = 0.6034 m
Explanation:
Given
L = 5 m
Wplank = 225 N
Wman = 522 N
d = 1.1 m
x = ?
We have to take sum of torques about the right support point. If the board is just about to tip, the normal force from the left support will be going to zero. So the only torques come from the weight of the plank and the weight of the man.
∑τ = 0 ⇒ τ₁ + τ₂ = 0
Torque come from the weight of the plank = τ₁
Torque come from the weight of the man = τ₂
⇒ τ₁ = + (5 - 1.1)*(225/5)*((5 - 1.1)/2) - (1.1)*(225/5)*((1.1)/2) = 315 N-m (counterclockwise)
⇒ τ₂ = Wman*x = 522 N*x (clockwise)
then
τ₁ + τ₂ = (315 N-m) + (- 522 N*x) = 0
⇒ x = 0.6034 m
Calculate the magnitude of the linear momen- tum for each of the following cases a) a proton with mass 1.67 × 10-27 kg mov- ing with a velocity of 6 × 106 m/s. Answer in units of kg · m/s.
The scientist is likely to be studying kinematics.
Kinematics is the branch of science, specifically physics, which is concerned with the motion of objects without reference to the forces that induce this motion. An example of kinematics is studying the change in velocity of an object over time or the distance covered by an object in a specified amount of time.