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inessss [21]
3 years ago
11

you and your friend left a bus terminal at the same time and traveled in opposite directions. Your bus was in heavy traffic and

had to travel 20 miles per hour slower than your freind's bus. After 3 hours, the buses were 270 miles apart. How fast was each bus going?
Physics
2 answers:
Luda [366]3 years ago
7 0

Answer:

speed of the two vehicle are 55 mph and 35 mph

Explanation:

given,

speed of friends vehicle = x mph

speed of your vehicle = (x - 20) mph

when both travel in opposite direction

distance between the two buses = 270 miles

distance = speed × time

270 = 3(x) + 3(x-20)                    

90 = 2 x -20                        

x = 55 mph                    

now, speed of other vehicle is (55-20) = 35 mph

hence, speed of the two vehicle are 55 mph and 35 mph

STALIN [3.7K]3 years ago
6 0

Answer:

The rate at which bus 1 is going is 55 mph

The rate at which bus 1 is going is 35 mph

Explanation:

As per the question:

Suppose, the distance traveled by Bus 1 be 'd' at the rate R after a time, t = 3h

Thus  

Suppose, the distance traveled by Bus 1 be 'd'' at the rate, R'20 mph slower than the rate of Bus 1 after the same time.

R' = R - 20

The distance is given as the product of rate and time:

d = Rt         (1)

Now, the total distance given is 270 miles:

d + d' = 270

Now, using eqn (1):

Rt + R't = 270

3(R + R - 20) = 270

6R = 270 + 60

R = 55 mph

R' = R - 20 = 55 - 20 = 35 mph

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A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Reception devices pick up the variation in the electric field vector of the electromagnetic wave sent out by the satellite. Given the satellite specifications listed in the problem introduction, what is the amplitude E0 of the electric field vector of the satellite broadcast as measured at the surface of the earth? Use ϵ0=8.85×10^−12C/(V⋅m) for the permittivity of space and c=3.00×10^8m/s for the speed of light.

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The electric field vector of the satellite broadcast as measured at the surface of the earth is  E_o = 6.995 *10^{-6} \ V/m

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From the question we are told that

     The height of the satellite is  r  = 35000 \ km  =  3.5*10^{7} \ m

      The power output of the satellite is P  = 1 \ KW  =  1000 \ W

       

Generally the intensity of the electromagnetic radiation of the satellite at the surface of the earth is  mathematically represented as  

     I  =  \frac{P}{4 \pi r^2}

substituting values

      I  =  \frac{1000}{4 * 3.142 (3.5*10^{7})^2}

      I  = 6.495*10^{-14} \  W/m^2

This intensity of the electromagnetic radiation of the satellite at the surface of the earth can also be   mathematically represented as  

          I  =  c * \epsilon_o * E_o^2

Where E_o is the amplitude of the electric field vector of the satellite broadcast so

         E_o =  \sqrt{\frac{2 * I}{c * \epsilon _o} }

substituting values

          E_o =  \sqrt{\frac{2 * 6.495 *10^{-14}}{3.0 *10^{8} * 8.85*10^{-12}} }

           E_o = 6.995 *10^{-6} \ V/m

 

   

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