To be able to determine the original speed of the car, we use kinematic equations to relate the acceleration, distance and the original speed of the car moving.
First, we manipulate the one of the kinematic equations
v^2 = v0^2 + 2 (a) (x) where v = 0 since the car stopped
Writing the equation in such a way that the initial velocity or v0 is written on one side of the equation,
<span>we get v0 = sqrt (2(a)(x))
Substituting the known values,
v0 = sqrt(2(3.50)(30.0))
v0 = 14.49 m/s
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Therefore, before stopping the car the original speed of the car would be 14.49 m/s
Answer:
Explanation:
The general consensus is that it's more “natural” to define distance (meter) and time (second) and as base units, and derive velocity a the ratio between them. ... The general consensus is that it's more “natural” to define distance (meter) and time (second) and as base units, and derive velocity a the ratio between them.
Answer:
<em>The cantaloupe has a speed of 117.6 m/s</em>
Explanation:
<u>Free Fall Motion</u>
It occurs when an object falls under the sole influence of gravity. Any object that is being acted upon solely by the force of gravity is said to be in a state of free fall. Free-falling objects do not face air resistance.
If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is
.
The final velocity of a free-falling object after a time t is given by:
vf=g.t
The cantaloupe has been dropped from rest. We are required to find the speed after t=12 seconds.
Calculate the final speed:
vf=9.8 * 12 = 117.6 m/s
The cantaloupe has a speed of 117.6 m/s
Answer:
Explanation:
change in the volume of the gas = 5.55 - 1.22
= 4.33 X 10⁻³ m³
external pressure ( constant ) P = 1 x 10⁵ Pa
work done on the gas
=external pressure x change in volume
= 10⁵ x 4.33 X 10⁻³
=4.33 x 10²
433 J
Using the formula
Q = ΔE + W , Q is heat added , ΔE is change in internal energy , W is work done by the gas
Given
Q = - 124 J ( heat is released so negative )
W = - 433 J . ( work done by gas is negative, because it is done on gas )
- 124 = ΔE - 433
ΔE = 433 - 124
= 309 J
There is increase of 309 J in the internal energy of the gas.
Answer:
If the positive object is touched to the electroscope, then the leaves would diverge because of the excess positive charge on the electroscope.
If then a negative object is brought near the electroscope, the excess positive charge on the leaves would be attracted to the external terminals of the electroscope and the leaves would again lose the distance of separation and be close to one another.