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s2008m [1.1K]
4 years ago
5

A car moves at a speed of 40 miles per hour for half an hour and then at 60 miles per hour for two hours. What is the average sp

eed, measured in miles per hour, for the entire trip?
Physics
1 answer:
katovenus [111]4 years ago
7 0

Answer:

The average speed for the entire trip is 56 miles/hour.

Explanation:

It is given that,

Initially, the car moves with a speed of, v₁ = 40 miles/hour

Initial time, t₁ = 0.5 hours

Distance covered during time t₁,

d_1=v_1t_1

d_1=40\times 0.5

d_1=20\ miles

Then it travels with a speed of, v₂ = 60 miles/hour

Then time taken, t₂ = 2 hours

Distance covered during time t₂,

d_2=v_2t_2

d_2=60\times 2

d_2=120\ miles

The average speed of the car can be calculated as :

v=\dfrac{d_1+d_2}{t_1+t_2}

v=\dfrac{120+20}{0.5+2}

v = 56 miles/hour

So, the average speed of the car for the entire trip is 56 miles/hour. Hence, this is the required solution.  

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Ivanshal [37]

Part 1)

here we know that supply took 10 s to reach the ground

so here we will have

y = \frac{1}{2}gt^2

y = \frac{1}{2}\times 9.8 \times 10^2

y = \frac{1}{2}\times 9.8 \times 100

y = 490 m

Part 2)

Here all the supply covered horizontal distance of 650 m in 10 s interval of time

so here we can say

speed = \frac{distance}{time}

v = \frac{650}{10}

v = 65 m/s


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An object is projected horizontally at 14.1 m/s from the top of a 195.0 meter cliff.
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7 0
3 years ago
Which of these black holes exerts the weakest tidal force on an object near its event horizon? Which of these black holes exerts
ratelena [41]

Answer:

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Explanation:

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6 0
3 years ago
Earth rotates on its axis once every 24 hours, so that objects on its surface execute uniform circular motion about the axis wit
cestrela7 [59]

Answer:

v=1667.9km/h

a_{cp}=436.6km/h^2

Explanation:

The speed is the distance traveled divided by the time taken. The distance traveled in 24hs while standing on the equator is the circumference of the Earth C=2\pi R, where R=6371km is the radius of the Earth.

We have then:

v=\frac{C}{t}=\frac{2\pi R}{t}=\frac{2\pi (6371km)}{(24h)}=1667.9km/h

And then we use the centripetal acceleration formula:

a_{cp}=\frac{v^2}{R}=\frac{(1667.9km/h)^2}{(6371km)}=436.6km/h^2

6 0
3 years ago
An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p
igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

Explanation:

Given:

Distance from the wire to the field point r = 2.83 \times 10^{-2} m

Speed of electron v = 35.5 \%c

Current I = 17.7 A

For finding the acceleration,

First find the magnetic field due to wire,

  B = \frac{\mu _{o}I }{2\pi r }

Where \mu_{o} = 4\pi   \times 10^{-7}

  B = \frac{4\pi \times 10^{-7}  \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }

  B = 12.50 \times 10^{-5} T

The magnetic force exerted on the electron passing through straight wire,

  F = qvB  

  F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}

  F = 21.3 \times 10^{-16} N

From the newton's second law

  F = ma

Where m = mass of electron = 9.1 \times 10^{-31} kg

So acceleration is given by,

   a = \frac{F}{m}

   a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }

   a = 2.34 \times 10^{15} \frac{m}{s^{2} }

Therefore, the magnitude of electron acceleration is 2.34 \times 10^{15} \frac{m}{s^{2} }

7 0
3 years ago
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