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3 years ago

11

A beach ball moving with the speed of 1.27 m/s rows of a pier and hits the water 0.75 m from the end the pier how high is the

pier above the water

1 answer:

Vera_Pavlovna [14]3 years ago

8 0

Answer:

75.8

Explanation:

because just divide 1.27 into 0.75 and there's your answer

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NASA developed a reflective foil barrier to help shield spacecraft from heat transfer in space. These reflective barriers are no

morpeh [17]

Answer: Radiation

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3 years ago

How long does it take anya to cover the distance of 5.00 miles ?

sammy [17]

Here is the full information about the question. <span>Ilya and Anya each can run at a speed of 8.50mph and walk at a speed of 3.50 mph . They set off together on a route of length 5.00 miles . Anya walks half of the distance and runs the other half, while Ilya walks half of the time and runs the other half.
the calculation would be:
</span><span>
t = d / s </span>
<span>t = 2.5 (half of the total distance) / 8.5 (speed of running) </span>
<span>This is .294 hours which is about 1058s... </span>

<span>for the walking part... </span>

<span>t = d / s </span>
<span>t = 2.5 / 3.5 </span>
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3 years ago

How can scientists study the inner structure of an atom?

Arturiano [62]

A microscope is one way

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3 years ago

The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh

Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

$v^2=v^2_o-2ax$

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

$\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}$

Next, consider that the formula for the force is:

$F=ma$

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

$F=(820kg)(0.66\frac{m}{s^2})=542.20N$

Hence, the required force to stop the car is 542.20N

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1 year ago

The angular momentum of an electron in each of its orbits must be some multiple of:

Nastasia [14]

The Bohr's proposal for the angular momentum of an electron in Bohr's model of the hydrogen atom is:

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3 years ago