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sveta [45]
3 years ago
12

A train leaves the station heading south on the tracks. It takes 5 seconds to reach 50 miles per hour. It completes the entire 1

00 mile trip in 2 hours. Calculate the trains average speed and velocity over the two hour trip. Show your work. Identify each of the measurements are a ascalr or vector.
Physics
2 answers:
Musya8 [376]3 years ago
8 0
Well, in order to figure out the answer is to divide until you figure out how many miles they went per second. If it takes 5 seconds to reach 50 miles per hour it took 10 seconds per every 10 miles meaning each mile took 1 second. (Not actually possible but the answer) So, If it finished a 100 mile trip in 2 hours it took an hour for 50 miles. If it took 1 hour for 50 miles divide 60/50 which gets you 1.2 so it took 1.2 miles per minute meaning the car went 120 miles per hour I believe. I hope this helps :)
PilotLPTM [1.2K]3 years ago
4 0

Answer:

Average speed is 50 mph

Average velocity is -50 mph . It is negative since it is towards south.

Explanation:

Distance is a scalar quantity and velocity is a vector. So for a vector, both magnitude and direction should be specified.

Average speed is the total distance over total time taken . So it is 100 miles over 2 hours which gives 50 mph.

Average velocity has the same magnitude and is also 100miles over 2 hours but it is negative since velocity is a vector and it is directed towards south.

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anyanavicka [17]

Answer:

1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

Explanation:

According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.

As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :

Q₁ = ∫ ρ dV

Here dV is the volume element of sphere of radius r.

Q₁ = ρ x 4π x ∫ r² dr

The limit of integration is from 0 to r as r is less than R.

Q₁ = (4π x ρ x r³ )/3

But volume charge density, ρ = \frac{3Q}{4\pi R^{3} }

So, Q_{1} = \frac{Qr^{3} }{R^{3} }

Applying Gauss law of electrostatics ;

∫ E ds = Q₁/ε₀

Here E is electric field inside the sphere and ds is surface element of sphere of radius r.

Substitute the value of Q₁ in the above equation. Hence,

E x 4πr² = ( Q x r³) / ( R³ x ε₀ )

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3 years ago
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Answer:

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Explanation:

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Answer:

B) Pressure on the scale, not registered as weight.

Explanation:

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Answer:

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Explanation:

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