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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Answer:
A horse is running at 12m/s accelerated to 38m/s in 10 seconds. What is the horses acceleration.
2.6m/s^2
Answer: Only Tech B is correct.
Explanation:
First, tech A is wrong.
The circuits that can be compared with links in a chain are the series circuit, and it can be related to the links in a chain because if one of the elements breaks, the current can not flow furthermore (because the elements in the circuit are connected in series) while in a parallel circuit if one of the branches breaks, the current still can flow by other branches.
Also in a parallel circuit, the sum of the currents of each path is equal to the current that comes from the source, so Tech B is correct, the total current is equal to the sum of the currents flowing in each branch of the circuit.
Answer:
The minimum inductance needed is 2.78 H
Explanation:
Given;
frequency of the AC, f = 26.5 Hz
the root mean square voltage in the circuit,
= 41.2 V
the maximum current in the circuit, I₀ = 126 mA
The root mean square current is given by;

The inductive reactance is given by;

The minimum inductance needed is given by;

Therefore, the minimum inductance needed is 2.78 H
Answer:
v = 384km/min
Explanation:
In order to calculate the speed of the Hubble space telescope, you first calculate the distance that Hubble travels for one orbit.
You know that 37000 times the orbit of Hubble are 1,280,000,000 km. Then, for one orbit you have:

You know that one orbit is completed by Hubble on 90 min. You use the following formula to calculate the speed:

hence, the speed of the Hubble is approximately 384km/min