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navik [9.2K]
3 years ago
10

Two families meet at the park at 10:00am. Each family uses a different way of transportation to get there. The Gonzalez family l

eaves at 9:00am and drives 90 km on a highway. The Rivera leave at 9:30am and ride the train 30km. What is the average speed for each family’s trip? Which family travels at the faster speed?
Physics
1 answer:
pav-90 [236]3 years ago
5 0

Firstly, we need to make the times and distance equal to compare.

1hr=60mins
Gonz drove 90km in 60mins,so they would drive 45km in 30 mins dividing it by 2.
Because the Rivs had gone 30km in 30mins, Gonz were faster.

For the adverage speed, we should first add the speed of each family in mph , so 90+60, which equals 150 and divide that by 2 because there are 2 speeds so the average speed is 75mph.

P.S English is my second language so tell me if you don't understand something and I will try and explain.
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Explanation:

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a = 3.79 x 10^14m/s^2

  • From F = ma
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  • ma = qE

E = ma /q = 9.11 x 10^-31 x 3.79 x 10^14 / 1.6 x 10^-19

E = magnitude of this electric field. = 2156.3N/C

b) Find the speed of the electron when it strikes the second plate ; V^2 = 2as

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Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

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What happens to the coefficient of friction when the weight is increased? Why is this?
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Calculate the kinetic energy that the earth has because of (a) its rotation about its own axis and (b) its motion around the sun
Dominik [7]

Answer:

a. K_{Axis}=2.574x10^{29}J

b. K_{Orbit}=2.6577x10^{33}J

Explanation:

K_{Axis}=\frac{1}{2}I*w^2

I_{Sphere}=\frac{2}{5}*m*r^2

w=\frac{2\pi }{T} , T=24hrs*\frac{3600s}{1hr} =86400s

radius earth = 6371 km

mass earth = 5,972*10^24 kg

a.

K_{Axis}=\frac{1}{2}*\frac{2}{5}*m*r^2*(\frac{2\pi}{T})^2

K_{axis}=\frac{4\pi^2}{5}*5.98x10^{24}kg*(6.38x10^6m)^2*(\frac{1}{86400s})^2

K_{Axis}=2.574x10^{29}J

b.

T=1year*\frac{365day}{1year}*\frac{24hr}{1day}*\frac{3600s}{1hr}=31536000s

K_{Orbit}=\frac{1}{2}*I*w

I=m*r^2

K_{Orbit}=\frac{1}{2}*m*r^2*(\frac{2\pi}{T})^2

K_{Orbit}=\frac{4\pi^2}{5}*5.98x10^{24}*6.38x10^6m*(\frac{1}{31.536x10^6s})^2

K_{Orbit}=2.6577x10^{33}J

4 0
4 years ago
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