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erica [24]
2 years ago
11

A school bus moves at speed of 35mi/he for 20miles. How long will it take for the bus to get to school

Physics
1 answer:
lara [203]2 years ago
8 0

Answer:

Time, t = 0.57 hours

Explanation:

It is given that,

Speed of the school bus, v = 35 mi/hr

Distance covered by the bus, d = 20 miles

We need to find the time taken by the bus to get to school. Time taken by the bus is given by :

t = 0.57 hours

So, the time taken by the bus to reach school is 0.57 hours. Hence, this is the required solution.

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Explanation:

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3 years ago
The diver uses compressed air to breathe under water. 1700 litres of air from the atmosphere is compressed into a 12-litre gas c
ivann1987 [24]

Answer:

The answer is below

Explanation:

A diver works in the sea on a day when the atmospheric pressure is 101 kPa. The diver uses compressed air to breathe under water. 1700 litres of air from the atmosphere is compressed into a 12-litre gas cylinder. The compressed air quickly cools to its original temperature. Calculate the pressure of the air in the cylinder.

Solution:

Boyles law states that the volume of a given gas is inversely proportional to the pressure exerted by the gas, provided that the temperature is constant.

That is:

P ∝ 1/V;   PV = constant

P₁V₁ = P₂V₂

Given that P₁ = initial pressure = 101 kPa, V₁ = initial volume = 1700 L, P₂ = cylinder pressure, V₂ = cylinder volume = 12 L. Hence:

P₁V₁ = P₂V₂

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6 0
2 years ago
A solid conducting sphere with radius R carries a positive total charge Q. The sphere is surrounded by an insulating shell with
Illusion [34]

Answer:

Explanation:

Volume of the insulating shell is,

V_{shell}=\frac{4}{3}\pi(R^3_2-R^3_1)

Charge density of the shell is,

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi(R^3_2-R^3_1)}

Here, R_2 =2R, R_1 =R \,and\, Q_{shell} =-Q

\rho=\frac{Q_{shell}}{\frac{4}{3}\pi((2R)^3-R^3)}=\frac{-3Q}{28\piR^3}

B)

The electric field is E=\frac{1}{4\pi\epsilon_0}\frac{Qr}{R^3}

For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.

C)

For R <r <2R According to gauss law

E(4\pi r^2)=\frac{Q}{\epsilon_0}+\frac{4\pi\rho}{3\epsilon_0}(r^3-R^3)

substitute \rho=\frac{-3Q}{28\piR^3}

E=\frac{2}{7\pi\epsilon_0}\frac{Q}{r^2}-\frac{Qr}{28\piR^3}

D)

The net charge enclosed for each r in this range is positive and the electric field is outward

E)

For r>2R

Charge enclosed is zero, so electric field is zero

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