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Ratling [72]
3 years ago
8

What problems do you think might arise from not having a clear definition of "high crimes and misdemeanors"?

Physics
1 answer:
damaskus [11]3 years ago
6 0

Answer:

People might misunderstand it and problems might happen because the audience will have different opinions.

Explanation:

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An _________ is a light ray moving toward a boundary.
Sveta_85 [38]

When A Light Ray Hits A Boundary,,,,, it is called refraction.

6 0
3 years ago
What is the thermal energy of an object?
nalin [4]
The answer would be A!
6 0
3 years ago
Read 2 more answers
a child in a tree house uses a rope attached to a basket to lift a 26 nn dog upward through a distance of 5.0 mm into the house.
agasfer [191]

Total work done is 0.13 Joules

<h3>What is work done ?</h3>

The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.

According to the given information

We need to find the work done

work done  = force × distance

We are given,

force  = 26 N

Distance = 0.0005 meter

hence ,

Work done  = 26 × 0.005  

                   = 0.13 Joules

Total work done is 0.13 Joules

To know more about Work done

brainly.com/question/13662169

#SPJ4

4 0
1 year ago
A crate with a mass of 110 kg glides through a space station with a speed of 4.0 m/s. An astronaut speeds it up by pushing on it
Darina [25.2K]

Answer:

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

Explanation:

<u>Given:  </u>

The crate has mass m = 110 kg and an initial speed vi = 4 m/s.  

<u>Solution  </u>

We are asked to determine the final speed of the crate. We could apply the steps for energy principle update form as next  

Ef=Ei+W                                                 (1)

Where Ef and Ei are the find and initial energies of the crate (system) respectively. While W is the work done by the astronaut (surrounding).  

The system has two kinds of energy, the kinetic energy which associated with its motion and the rest energy where it has zero speed. The summation of both energies called the particle energy. So, equation (1) will be in the form  

(Kf + mc^2) = (KJ+ mc^2)                       (2)  

Where m is the mass of crate, c is the speed of light which equals 3 x 10^8 m/s and the term mc^2 represents the energy at rest and the term K is the kinetic energy.  

In this case, the rest energy doesn't change so we can cancel the rest energy in both sides and substitute with the approximate expression of the kinetic energy of the crate at low speeds where K = 1/2 mv^2 and equation (2) will be in the form

(1/2mvf^2+mc^2)=(1/2mvi^2 +mc^2)+W

1/2mvf^2=1/2mvi^2+W                              (3)

Now we want to calculate the work done on the crate to complete our calculations. Work is the amount of energy transfer between a source of an applied force and the object that experiences this force and equals the force times the displacement of the object. Therefore, the total work done will be given by  

W = FΔr                                                      (4)  

Where F is the force applied by the astronaut and equals 190 N and Δr is the displacement of the crate and equals 6 m. Now we can plug our values for F and Δr to get the work done by the astronaut  

W = F Δr= (190N)(6 m) = 1140 J  

Now we can plug our values for vi, m and W into equation (3) to get the final speed of the crate  

1/2mvf^2=1/2mvi^2+W

vf=5.82 m/s

This is the final speed of the first push when the astronaut applies a positive work done. Then, in the second push, he applies a negative work done on the crate to slow down its speed. Hence, in this case, we could consider the initial speed of the second process to be the final speed of the first process. So,  

vi' = vf

In this case, we will apply equation (3) for the second process to be in the

1/2mvf^2=1/2mvi'^2+W'                                 (3*)

The force in the second process is F = 170 N and the displacement is 4 m. The force and the displacement are in the opposite direction, hence the work done is negative and will be calculated by  

W'= —F Δr = —(170N)(4 m)= —680J

Now we can plug our values for vi' , m and W' into equation (3*) to get the final speed of the crate  

1/2mvf'^2=1/2mvi'^2+W'

  vf'=4.50 m/s

The final speed of the crate after the astronaut push to slow it down is 4.50 m/s

7 0
3 years ago
A 0.0750kg teddy bear is dropped from a deck that is 3.00m above the ground. What will be the velocity of the teddy bear as it s
FrozenT [24]

The velocity of the teddy bear as it strikes the ground is 7.67 m/s.

<h3>Velocity of the teddy when it strikes the ground</h3>

The velocity of the teddy when it strikes the ground is calculated from principle of conservation of energy as shown below.

K.E(bottom) = P.E(top)

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

  • h is height of fall of the teddy
  • g is acceleration due to gravity

v = √(2 x 9.8 x 3)

v = 7.67 m/s

Thus, the velocity of the teddy bear as it strikes the ground is 7.67 m/s.

Learn more about conservation of energy here: brainly.com/question/166559

#SPJ1

4 0
1 year ago
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