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Lisa [10]
3 years ago
10

A 53.5-g Super Ball traveling at 27.5 m/s bounces off a brick wall and rebounds at 21.5 m/s. A high-speed camera records this ev

ent. If the ball is in contact with the wall for 4.05 ms, what is the magnitude of the average acceleration of the ball during this time interval
Physics
1 answer:
Tema [17]3 years ago
4 0

Answer:

<h3>12,098.77m/s²</h3>

Explanation:

Acceleration is the rate of change of velocity of a body. It is expressed as;

acceleration = change in velocity/time

acceleration a = v-u/t

Given

v = 27.5m/s

u = -21.5m/s (rebounds in the opposite direction)

time t = 4.05ms = 0.00405secs

Required

acceleration

Substitute the given values into the formula

a = 27.5-(-21.5)/0.00405

a = 27.5+21.5/0.00405

a = 49/0.00405

a = 12,098.77m/s²

Hence  the magnitude of the average acceleration of the ball during this time interval is 12,098.77m/s²

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Convert 3402kgm/s to 20000Newtons
oee [108]

The 3,402 has units of kg-m/s.  That's momentum.  The 20,000 has units of Newtons.  That's force.  Momentum and force are different physical things, and you can't convert them from one to the other.

The best I can do for you is something like this:

Let's say you have a moving object with 3,402 kg-m/s of momentum, and you want to STOP it completely.  You want to stand in front of it and push back on it, hard enough and for long enough to CHANGE its momentum from 3,402 kg-m/s to zero.

Also ... there's a limit to how hard you can push.  The most force you can exert is 20,000 Newtons.

The amount you'll change its momentum is called the <u><em>impulse</em></u> you give it.  The quantity of impulse is (force) x (length of time you push on it).

So you need to keep pushing it back for (T seconds) long enough so that

(20,000 Newtons of force) x (T seconds) = 3,402 kg-m/s of momentum .

Divide each side of that equation by (20,000 Newtons). Then it says:

(T seconds) = (3,402 kg-m/s) / (20,000 Newtons)

<em>T = 0.1701 second</em>

And that's how you provide just enough impulse to stop the flying object ... push on it with 20,000 Newtons of force for exactly 0.1701 second, and it loses all its momentum and falls out of the air onto the ground at your feet.

This story is the closest I can come to anything that looks like "convert"ing momentum into force.

3 0
3 years ago
Why do adults make bigger splashes when they jump into swimming pools than small children?
Semenov [28]

Answer:

it bc the adults are bigger then us kid so when they dip in the pool it makes bigger splashes

Explanation:

5 0
3 years ago
Will give brainlist!! Please help!!
Marrrta [24]

Answer:Electromagnetic Energy Example One

activity: cellphones

type of electromagnetic: radio waves

description: we all use our phones to make phone calls and to send a text!

Electromagnetic Energy Example two

activity: microwave

type of electromagnetic: microwave radiation

description: The microwave radiation is absorbed by water molecules in the food which converts to heat intern heats the food do to high levels of radiation being emitted into the food!  

Explanation:

i hope this helps you sorry if it doesn't

5 0
3 years ago
Calculate the average speed in metres per second from Glasgow to Edinburgh
mariarad [96]
This is the same question as the one previously but with more details, so I will just use my previous answer.

1800 to 1820 is 20 minutes.1830 to 1838 is 8 minutes.1840 to 1905 is 25 minutes.
The total time travelled is 20+8+25 = 53 minutes = 3180 seconds.
The distance between Glasgow and Edinburgh is 28 + 12 + 34 = 74 km = 74000 m.

So, the average speed is 74000m/3180s = 23.27 m/s (4 s.f.)
5 0
3 years ago
Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc
Crazy boy [7]

Answer:

Temperature at the exit = 267.3 C

Explanation:

For the steady energy flow through a control volume, the power output is given as

W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

Inlet area of the turbine = 60cm^{2}= 0.006m^{2}

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume v_{1}

as

v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg

m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec

for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

Hence, the temperature at the exit = 267.3 C

5 0
3 years ago
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