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KonstantinChe [14]
4 years ago
12

An open train car, with a mass of 2130 kg, coasts along a horizontal track at the speed 2.93 m/s. The car passes under a loading

chute and, as it does so, gravel falls vertically into it for 3.03 s at the rate of 465 kg/s. What is the car's speed v f after the loading is completed
Physics
1 answer:
crimeas [40]4 years ago
4 0

Answer:

The final speed after the loading is completed is 1.764 m/s

Explanation:

Given;

mass of open train car, m₁ = 2130 kg

initial speed of the car, u₁ = 2.93 m/s

loading rate of gravel = 465 kg/s

time of loading = 3.03 s

mass of the gravel, m₂ = 465 kg/s x 3.03 s = 1408.95 kg

From the principle of conservation of linear momentum

m₁u₁ + m₂u₂ = v(m₁+m₂)

where;

v is the final speed after the loading is completed

2130 x 2.93 + 0 = v (2130 + 1408.95)

6240.9 = v (3538.95)

v = 6240.9/3538.95

v = 1.764 m/s

Therefore, the final speed after the loading is completed is 1.764 m/s

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3 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
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Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

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3 years ago
A horizontal beam of light of intensity 25 W/m2 is sent through two polarizing sheets. The polarizing direction of the first mak
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Let the intensity of light as it passes from second polariser is I''.

According to the law of Malus

I'' = I' Cos^{2}\theta

Where, θ be the angle between the axis first polariser and the second polariser.

I'' = 12.5\times Cos^{2}15

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Answer:

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