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stiks02 [169]
2 years ago
6

Guyz wqy-bdet-mby join m.e f.o.r f.u.n​

Physics
1 answer:
sergey [27]2 years ago
4 0

Answer:

What?  

Explanation:

You might be interested in
Velocity is:
sleet_krkn [62]

Answer:

d

Explanation:

Solution:-

- The Quantity of theory of money states:

                      M * V = P * Y

Where,

           M = Money supply

           V = Velocity of money exchange

           P = The price level

           Y = Real GDP

- By re-arranging the formula and solving for "V" we have:

                     V = P*Y / M

- The expression on right hand side increases if exchange of dollars increases.

3 0
3 years ago
A typical land snail's speed is 12.2 meters per hour. How many miles will the snail travel in one day(24hrs)?
algol13
The snail will go <span>0.18193752 miles </span>
5 0
3 years ago
Earth's surface receives about twice as much energy from the atmosphere than from the sun as a result of ____.
Oxana [17]
I think the answer is “greenhouse effect”
5 0
3 years ago
A string attached to an airborne kite was maintained at an angle of 40.0 with the ground. if 120m of string was reeled in to ret
AleksAgata [21]
The hypotenuse is measured at 120 meters of string, and you need to solve for the leg of the triangle that is horizontal. The degree is 40, so use trigonometry to figure it out. 
Cosin (40) is equal to around .766 
Adjacent/Hypotenuse 
x/120 = cos40 
Answer: 91.92533. 
If you use 3 significant figures it should be 91.9 meters.
5 0
3 years ago
As a quality test of ball bearings, you drop bearings (small metal balls), with zero initial velocity, from a height of 1.94 m i
11Alexandr11 [23.1K]

Answer:

The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

Explanation:

Given that,

Height = 1.94  m

Bounced height = 1.48 m

Time interval t=14.86\times10^{-3}\ s

Velocity of the ball bearing just before hitting the steel plate

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

9.8\times1.94=\dfrac{1}{2}\times v^2

v=\sqrt{2\times9.8\times1.94}

v=6.166\ m/s

Negative as it is directed downwards

After bounce back,

We need to calculate the velocity

Using conservation of energy

mgh=\dfrac{1}{2}mv^2'

Put the value into the formula

9.8\times1.48=\dfrac{1}{2}\times v^2'

v'=\sqrt{2\times9.8\times1.48}

v'=5.38\ m/s

We need to calculate the average acceleration of the bearings while they are in contact with the plate

Using formula of acceleration

a=\dfrac{v-v'}{t}

Put the value into the formula

a=\dfrac{5.38-(-6.166)}{14.86\times10^{-3}}

a=776.98\ m/s^2

a=0.77\times10^{3}\ m/s^2

Hence,The average acceleration of the bearings is 0.77\times10^{3}\ m/s^2

6 0
3 years ago
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