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4 years ago

Which of the following is an example of converting kinetic energy into potential energy?

Physics

2 answers:

ratelena [41]4 years ago

7 0

a. dropping your backpack on the floor

Margarita [4]4 years ago

4 0

I believe that the answer to this is d. picking up your backpack to get out an assignment.

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A moon orbits a planet every 42 hours with a mean orbital radius of .002819 AU. The mass of the moon is 8.932 x 1022 kg. Using N

Pepsi [2]

Answer:

The mass of the planet is $1.9407\times10^{27}\ kg$

Explanation:

Given that,

Time period = 42 hours = 151200 sec

Orbital radius = 0.002819 AU = 421716397.5 m

Mass of moon $m=8.932\times10^{22}\ kg$

We need to calculate the mass of the planet

Using Kepler’s third law

$T^2\propto a^3$

$T^2=\dfrac{4\pi^2}{G(M+m)}\times a^3$

Where, a = orbital radius

T = time period

G = gravitational constant

M = mass of moon

m = mass of planet

Put the value into the formula

$(151200)^2=\dfrac{4\pi^2}{6.673\times10^{-11}(8.932\times10^{22}+m)}\times(421716397.5)^3$

$(8.932\times10^{22}+m)=\dfrac{4\pi^2}{6.673\times10^{-11}}\times\dfrac{(421716397.5)^3}{(151200)^2}$

$(8.932\times10^{22}+m)=1.94087\times10^{27}$

$m=1.94087\times10^{27}-8.932\times10^{22}$

$m=1.9407\times10^{27}\ kg$

Hence, The mass of the planet is $1.9407\times10^{27}\ kg$

8 0

4 years ago

What are distillation and evaporation most often used to separate in a science lab and why?

Alex777 [14]

C because I think that’s the answer

3 0

2 years ago

Read 2 more answers

What is the unit of density

maria [59]

Density = Mass/Volume. The units for density are grams per cubic centimeter or grams per milliliter

3 0

3 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

3 0

3 years ago

If a car takes 3500j of total energy to make 1200j of useful energy, how energy efficient is it

Luden [163]

Answer: 34%

Explanation: eff = (Eout/Ein) *100(for % form)

1200/3500 * 100 = 34%

4 0

3 years ago