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Vinvika [58]
4 years ago
12

a box has a momentum of 1.30 kg*m/s to the right. a 12.8 N force pushes it to the right for 0.218 s. what is the final momentum

of the box. PLEASE HELP
Physics
1 answer:
Rufina [12.5K]4 years ago
5 0

Answer:

4.09 kgm/s

Explanation:

Concept tested: Second Newton's law of motion

So what does the second Newton's law of motion state?

  • According to the second Newton's law of motion, the resultant force and the rate of change of momentum are directly proportional.
  • That is; F=\frac{(mVf-mVo)}{t}, where, mVf is the final momentum and mVo is the initial momentum.

In this case;

  • Force = 12.8 N
  • Initial momentum, mVo = 1.3 kgm/s
  • Time, t = 0.218 seconds

Therefore; replacing the values of the variables in the formula;

we get;

12.8N=\frac{(mVf-1.3kgm/s)}{0.218s}

(12.8N)(0.218s)=(mVf-1.3kgm/s)

2.7904+1.3=mVf

Solving for final momentum;

mVf =4.0904 kgm/s

      = 4.09 kgm/s

Therefore, the final momentum of the box is 4.09 kgm/s

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Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22
natka813 [3]

Answer:

2420 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 22.9 N

Angle (θ) = 35°

Distance (d) = 129 m

Workdone (Wd) =?

The work done can be obtained by using the following formula:

Wd = Fd × Cos θ

Wd = 22.9 × 129 × Cos 35

Wd = 22.9 × 129 × 0.8192

Wd ≈ 2420 J

Thus, the workdone is 2420 J.

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3 years ago
Calculate the lateral displacement of a ray of light incident at an angle of 30o on a parallel – sided block of glass of thickne
OLga [1]
The lateral displacement is I don’t know tbh I think 16.8
6 0
3 years ago
Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos
soldi70 [24.7K]
<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

           F₁ = 100 cos 20 i  +  100 sin 20 j

           F₁ = 93.97 i  +  34.20 j          

That is force 2 is 150 pound with x axis at 60°

           F₂ = 150 cos 60 i  +  150 sin 60 j

           F₂ = 75 i  +  129.90 j  

F₁ +  F₂ =  93.97 i  +  34.20 j + 75 i  +  129.90 j

F₁ +  F₂ =  168.97 i  +  164.10 j

\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0
3 years ago
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1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}
substituting the coordinates of the two charges, we get
d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m

2) Then, we can calculate the electrostatic force between the two charges q_1 and q_2, which is given by
F=k_e  \frac{q_1 q_2}{d^2}
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Substituting numbers, we get 
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Inessa [10]
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