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Karolina [17]
3 years ago
13

How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m? (The acceleration due to gravity is 9.8 m/s2.)

Physics
2 answers:
V125BC [204]3 years ago
7 0
<span>How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?
</span>Answer:

85 Joule (approx)

Explanation:

Potential energy at highest point

<span><span>P.E = mgh = 7.2 kg × 9.8 m/s2</span>×1.2 m≈85 Joule</span>

This is the kinetic energy required for dog to jump a height of 1.2 m.



hope this helps!

Tasya [4]3 years ago
6 0

Answer:

Kinetic energy of dog is 84.67 Joules

Explanation:

Given that,

Mass of the dog, m = 7.2 kg

We have to find the kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m. We know that the energy of a system remains conversed. At highest point there is only potential energy. So, while taking a vertical jump the potential energy gets converted to potential energy.

So, E_k=mgh

here, h = 1.2 m

E_k=7.2\times 9.8\times 1.2

E_k=84.67\ Joules

Hence, the kinetic energy needed to make a vertical jump is 84.67 Joules.

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Explanation:

<h2>Yes!</h2>

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What is the change in potential energy if the distance separating the electron and proton is increased to 1.0 nm?
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Answer:

Ep=-2.3*10^{-19}J

Explanation:

The change in potential energy can be expressed as:

Ep=K.\frac{q1.q2}{r}

where K is a constant with a value of 9*10^{9}\frac{N.m^{2}}{C^{2}}, q1 and q2 are the charges of the proton and the electron and r is the distance between them.

The charge for the proton is +1.6*10^{-19}C and the charge for the electron is -1.6*10^{-19}C.

Converting r=1.0nm to m:

1.0nm*\frac{1*10^{-9}m}{1.0nm}=1*10^{-9}m

Replacing values:

Ep=9*10^{9}\frac{N.m^{2}}{C^{2}}.\frac{(+1.6*10^{-19}C).(-1.6*10^{-19}C)}{1*10^{-9}m}

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2 years ago
The diagram below outlines the appeals path through a state court system.
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Answer:

B sorry if im wrong

Explanation:

4 0
3 years ago
Read 2 more answers
Calculate the approximate volume of a uranium nucleus, 23592u. (you can ignore the mass defect in this calculation and simply ta
tester [92]

Radius of nuclei is given by formula

R = R_oA^{1/3}

now we can say volume of the nuclei is given as

V = \frac{4}{3}\pi R_o^3* A

now the density is given as

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mass of nuclei = mass of neutron + mass of protons

m = z*m_p + (A- z)*m_n

m_p = m_n = 1.008u

m = A*1.008u

Now density is given as

\rho = \frac{A*1.008u}{\frac{4}{3}\pi R_0^3* A}

here we know that

R_0 = 1.2 fm

\rho = \frac{1.008u}{\frac{4}{3}\pi*(1.2*10^{-15})^3}

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3 years ago
A coat rack weighs 65.0 lbs when it is filled with winter coats and 40.0 lbs when it is empty. The base of the coat rack has an
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Answer:

0.056 psi more pressure is exerted by filled coat rack than an empty coat rack.

Explanation:

First we find the pressure exerted by the rack without coat. So, for that purpose, we use formula:

P₁ = F/A

where,

P₁ = Pressure exerted by empty rack = ?

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P₁ = 40 lb/452.4 in²

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Now, we calculate the pressure exerted by the rack along with the coat.

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ΔP = P₂ - P₁

ΔP = 0.144 psi - 0.088 psi

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