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Nuetrik [128]
2 years ago
11

Recall that impulse = momentum (FAt = Ap and that Ap is just mx v). How long (time) must a group of people pull with a force of

600N on a stalled 1500kg car to give it a velocity of 1.5m/s? Report your answer to the nearest hundredth (X.XX)​
Physics
1 answer:
elixir [45]2 years ago
6 0

The time taken for the group of people to pull the car, giving it a velocity of 1.5 m/s is 3.75 s

Momentum is simply defined as the product of mass and velocity i.e

Momentum = mass × velocity

To answer the question given, we'll begin by calculating the change in momentum. This can be obtained as follow:

Mass = 1500 Kg

Initial velocity (u) = 0 m/s

Final velocity (v) = 1.5 m/s

<h3>Change in momentum =? </h3>

Change in momentum = m(v – u)

Change in momentum = 1500 (1.5 – 0)

Change in momentum = 1500 × 1.5

<h3>Change in momentum = 2250 Kg•m/s</h3>

Finally, we shall determine the time

Change in momentum = 2250 Kg•m/s

Force (F) = 600 N

<h3>Time (t) =? </h3>

Impulse = Ft = change in momentum

FT = change in momentum

600 × t = 2250

Divide both side by 600

t = 2250 / 600

<h3>t = 3.75 s</h3>

Thus, the time required is 3.75 s

Learn more on momentum and impulse: brainly.com/question/14486244

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1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x
allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }

With (\epsilon_{0}) been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    }

Explanation:

(A) Considering a uniform linear density λ_{0} on the ring, then:

dQ=\lambda dl (1)⇒Q=\lambda_{0} 2\pi R(2)⇒\lambda_{0}=\frac{Q}{2\pi R}(3)

Applying the technique of charge integration for finite charges:

V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r'  }} \, dQ(4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

r'=\sqrt{R^{2} +Z^{2}}(5)

Using the expressions (1),(4) and (5) you obtain:

V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}}  }} \, d\phi

Integrating results:

V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}}  }   (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

|E| =-\nabla V (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

|E| =-\frac{dV(z)}{dz} (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} }    } (S_b)

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What is called gravitational potential energy?
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Answer:

Potential gravitational energy is the energy that the body has due to the Earth's gravitational attraction. In this way, the potential gravitational energy depends on the position of the body in relation to a reference level.

Explanation:

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Something that can not be used up or depleted​
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Answer:

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Explanation:

as it is in the air it can't be depleted or used up

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Read 2 more answers
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