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Norma-Jean [14]
3 years ago
5

A horizontal force of 150 N is used to push a 38.0 kg packing crate a distance of 6.85 m on a rough horizontal surface. If the c

rate moves at constant speed, find each of the following.
a) find the work done by the 150N force:
b) find the coefficient of kinetic energy between the crate and rough surface
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer:

0.40

Explanation:

a) Work done is expressed as

W = Force × distance

Given.

Force = 150N

Distance = 6.85m

Workdone = 150×6.85

Work done = 1027.5Joules

b) According to Newton's second law

\sumFx = ma

Fm - Ff = ma

Since speed is constant, acceleration us zero

Fm - Ff = 0

Fm = Ff = nR

Fm is the moving force

Ff is the frictional force

n is the coefficient of kinetic energy between the crate and rough surface

R is the reaction

From the formula;

Fm = nR

n = Fm/R

n = Fm/mg

n = 150/38(9.8)

n = 150/372.4

n = 0.40

Hence the coefficient of kinetic energy between the crate and rough surface is 0.40

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4 0
3 years ago
The Hubble Space Telescope has a mass of 1.16*10^ 4 kg and orbits the Earth at an altitude of 5.68 * 10 ^ 5 above Earth's surfac
andrezito [222]

Answer:

E=8.13\times 10^{12}\ J

Explanation:

Given that,

The mass of a Hubble Space Telescope, m_1=1.16\times 10^4\ kg

It orbits the Earth at an altitude of 5.68\times 10^5\ m

We need to find the potential energy the telescope at this location. The formula for potential energy is given by :

E=\dfrac{Gm_1m_e}{r}

Where

m_e is the mass of Earth

Put all the values,

E=\dfrac{6.67\times 10^{-11}\times 1.16\times 10^4\times 5.97\times 10^{24}}{5.68\times 10^5}\\\\E=8.13\times 10^{12}\ J

So, the potential energy of the telescope is 8.13\times 10^{12}\ J.

5 0
2 years ago
Which elements will bond ionically with barium such that the formula would be written as BaX2? A) Nitrogen, chlorine, and sodium
erma4kov [3.2K]

The answer would be:

B. Chlorine, iodine and Fluorine

Barium has 2 valence electrons. To satisfy the BaX₂ , this would mean that Barium will need to give one of each of its electrons. The elements that need 1 electron would be those that have 7 valence electrons to complete the octet. These elements would fall in group 7 or halogens. Chlorine, iodine and fluorine are all in Group 7, so this would be the best choice.

7 0
3 years ago
Read 2 more answers
Is it proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinde
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Answer:

No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.

Explanation:

A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.

Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:

* When the diameter and length are comparable (i.e have the same measurement)

When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.

Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.

8 0
2 years ago
One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
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