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trasher [3.6K]
4 years ago
6

Constanza is on a commuter train between Richville and Shoptown. The train takes 35 minutes to cover the distance between the tw

o towns. The train's average speed is 55 kilometers per hour. Use the the concepts of displacement, velocity, and acceleration to describe the train's motion. Calculate the parameters of motion where possible.
Physics
1 answer:
just olya [345]4 years ago
5 0

Answer:

1.) Displacement = 32.08m

2.) Velocity = 55km/h

3.) Acceleration = 94.29 km/h^2

Explanation:

Given that the train takes 35 minutes to cover the distance between the two towns. The train's average speed is 55 kilometers per hour. That is,

Time t = 35 minute

Convert it to hours by dividing it by 60

35/60 = 7/12 hours

Speed = 55 km/h

The displacement of the car is:

Displacement = 55 × 7/12 = 32.08 m

The velocity of the car will be 55 km/h

The acceleration of the car will be:

Acceleration = velocity/time

Substitute velocity and time into the formula

Acceleration = 55 ÷ 7/12

Acceleration = 94.29 km/h^2

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A car travels 92 miles in 2 hours. What is the car's AVERAGE SPEED?
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3 years ago
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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