The second diver have to leap to make a competitive splash by 4.08 m high.
<h3>What is potential energy?</h3>
The energy by virtue of its position is called the potential energy.
PE = mgh
where, g = 9.81 m/s²
Given is the diver jumps from a 3.00-m platform. one diver has a mass of 136 kg and simply steps off the platform. another diver has a mass of 100 kg and leaps upward from the platform.
The potential energy of the first diver must be equal to the second diver.
P.E₁ = P.E₂
m₁gh₁ = m₂gh₂
Substitute the vales, we have
136 x 3 = 100 x h₂
h₂ = ₂4.08 m
Thus, the second diver need to leap by 4.08 m high.
Learn more about potential energy.
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Multiply by (1000 meters / 1 km).
Then multiply by (1 hour / 3600 seconds).
Both of those fractions are equal to ' 1 ', because the top
and bottom numbers are equal, so the multiplications
won't change the VALUE of the 72 km/hr. They'll only
change the units.
(72 km/hour) · (1000 meters / 1 km) · (1 hour / 3600 seconds)
= (72 · 1000 / 3600) (km·meter·hour / hour·km·second)
= 20 meter/second
<span>this may help you
As far as the field goes, the two charges opposite each other cancel!
So E = kQ / d² = k * Q / (d/√2)² = 2*k*Q / d² ◄
and since k = 8.99e9N·m²/C²,
E = 1.789e10N·m²/C² * Q / d² </span>
Answer:
a) t1 = v0/a0
b) t2 = v0/a0
c) v0^2/a0
Explanation:
A)
How much time does it take for the car to come to a full stop? Express your answer in terms of v0 and a0
Vf = 0
Vf = v0 - a0*t
0 = v0 - a0*t
a0*t = v0
t1 = v0/a0
B)
How much time does it take for the car to accelerate from the full stop to its original cruising speed? Express your answer in terms of v0 and a0.
at this point
U = 0
v0 = u + a0*t
v0 = 0 + a0*t
v0 = a0*t
t2 = v0/a0
C)
The train does not stop at the stoplight. How far behind the train is the car when the car reaches its original speed v0 again? Express the separation distance in terms of v0 and a0 . Your answer should be positive.
t1 = t2 = t
Distance covered by the train = v0 (2t) = 2v0t
and we know t = v0/a0
so distanced covered = 2v0 (v0/a0) = (2v0^2)/a0
now distance covered by car before coming to full stop
Vf2 = v0^2- 2a0s1
2a0s1 = v0^2
s1 = v0^2 / 2a0
After the full stop;
V0^2 = 2a0s2
s2 = v0^2/2a0
Snet = 2v0^2 /2a0 = v0^2/a0
Now the separation between train and car
= (2v0^2)/a0 - v0^2/a0
= v0^2/a0
When placing the piece of aluminium in water, the level of water will rise by an amount equal to the volume of the piece of aluminum.
Therefore, we need to find the volume of that piece.
Density can be calculated using the following rule:
Density = mass / volume
Therefore:
volume = mass / density
we are given that:
the density = 2.7 g / cm^3
the mass = 16 grams
Substitute in the equation to get the volume of the piece of aluminum as follows:
volume = 16 / 2.7 = 5.9259 cm^3
Since the water level will rise to an amount equal to the volume of aluminum, therefore, the water level will rise by 5.9259 cm^3
