The international governing body for badminton is the

Part A Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.7 m whose uniform depth is 1.8 m . Express y

Sphinxa [80]

Answer:

17.66 kPa

Explanation:

The volume of water in the swimming pool is the product of its dimensions

V = 30 * 8.7 * 1.8 = 469.8 cubic meters

Let water density $\rho = 1000 kg/m^3$ , and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool

$W = mg = \rho V g = 1000 * 469.8 * 9.81 = 4608738 N$

The area of the bottom

A = 30 * 8.7 = 261 square meters

Therefore the pressure is its force over unit area

$P = F/A = 4608738 / 261 = 17658 N/m^2$ or 17.66 kPa

7 0

3 years ago

The length of a wire 2.00 m is measured as 2.02m. What is the percentage error in the measurement?

n200080 [17]

Answer:

1%

Explanation:

Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.

$Percent Error=\frac{V_{measured}- V_{true} } {V_{true}} *100$

The measured value is 2.02 meters and the actual value is 2.00 meters.

$V_{measured}=2.02\\\\V_{true}=2.00$

$Percent Error=\frac{2.02-2.00}{2.00} *100$

First, evaluate the fraction. Subtract 2.00 from 2.02

$Percent Error=\frac{0.02}{2.00}*100$

Next, divide 0.02 by 2.00

$PercentError=0.01 *100$

Finally, multiply 0.01 and 100.

$Percent Error=1\\Percent Error= 1 \%$

The percent error is 1%.

6 0

3 years ago

What is the mass of a large ship that has a momentum of 1.60×109kg·m/s, when the ship is moving at a speed of 48.0 km/h? (b) Com

erastova [34]

a) The mass of the ship is $1.2\cdot 10^8 kg$

b) The ship has a larger momentum than the shell

Explanation:

a)

The momentum of an object is given by:

$p=mv$

where

m is the mass of the object

v is its velocity

For the ship in this problem, we have

$p=1.60\cdot 10^9 kg m/s$ is the momentum

$v=48.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=13.3 m/s$ is the velocity

Solving for m, we find the mass of the ship:

$m=\frac{p}{v}=\frac{1.60\cdot 10^9}{13.3}=1.2\cdot 10^8 kg$

b)

The momentum of the artillery shell is given by

$p=mv$

where

m is its mass

v is its velocity

For the shell in this problem,

m = 1100 kg

v = 1200 m/s

Substituting,

$p=(1100)(1200)=1.32\cdot 10^6 kg m/s$

So, we see that the ship has a larger momentum.

Learn more about momentum:

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brainly.com/question/6573742

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#LearnwithBrainly

6 0

3 years ago

It took Lilly 7 seconds to ride her bike 35 meters to the end of the block.What was her speed?

Zolol [24]

<h2><em>her average speed was 5 meter p/ second </em></h2><h2><em> 18 kph</em></h2><h2><em> HOPE IT HELPS (◕‿◕✿) </em></h2><h2><em> SMILE!!</em></h2>

4 0

2 years ago

Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1, the ball bounces

Kitty [74]

Answer:

Explanation:

1. The balls are identical in all sense, which means that if they are dropped from the same height, they should posses the same kinetic energy just before they collide with either the concrete floor or the stretchy rubber. Also, since they reach the same height when they bounced of the concrete floor or the piece of stretchy rubber, it means that they posses the same amount of kinetic energy at this point. Since their kinetic energy at these two points are the same, and they have the same masses, then this means that their momenta at these two instances will also be equal. Since all these is true, then the change in the momentum of the balls between the instance just before hitting the concrete floor or the stretchy rubber material and the instant the ball just leave the floor or the stretchy material is the same for both.

2. The ball that falls on the concrete will experience the greatest force, since the time of impact is small, when compared to the time spent by the other ball in contact with the stretchy rubber material; which will stretch, thereby extending the time spent in contact between them.

7 0

3 years ago