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GrogVix [38]
3 years ago
10

While traveling, John let destination A at 9 am. Driving until 7 pm, he travelled a total of 500 miles. The next day, John left

at 9 am and drove until 5 pm, covering 270 miles. What was the average speed for both days, while driving
Physics
2 answers:
guapka [62]3 years ago
8 0

Answer:

The average speed for both days, while driving is 42.77 miles/hr.

Explanation:

Given that,

Distance first day = 500 miles

Distance second day = 270 miles

According to question,

The distance covered by the john in 10 hours = 500 miles

Next day ,

The distance covered by the john in 8 hours = 270 miles

We need to calculate the average speed for both days

Using formula of the average speed

v_{avg}=\dfrac{D}{T}

Where, D = total distance

T = total time

Put the value into the formula

v_{avg}=\dfrac{500+270}{10+8}

v_{avg}=\dfrac{770}{18}

v_{avg}=42.77\ miles/hr

Hence, The average speed for both days, while driving is 42.77 miles/hr.

beks73 [17]3 years ago
5 0

18 hours 770

770/18 average speed.

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A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in
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Answer:

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Explanation:

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The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N

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A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
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Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

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z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

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Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

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Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
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