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ivolga24 [154]
3 years ago
12

5. How much does a 20 m x 10 m x 8 m swimming pool filled with water weigh? Assume that water has a density of 62 kg/m².​

Physics
2 answers:
Licemer1 [7]3 years ago
6 0

Answer:

fxhehjejejkwkwkkwkwkqklwkjsnenrnkeke

scoray [572]3 years ago
3 0

Answer:

99,200kg

Explanation:

density= mass/volume

given d-62kg/m*3

v=20×10×8. =1600

m=x

1600×62=99,200kg

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A student at the top of building of height h throws one ball upward with the initial speed V and then throws a second ball downw
Vlad1618 [11]

Answer:

They are the same (assuming there is no air friction)

Explanation:

Take a look at the picture.

When the first ball (the one thrown upward) gets to the point marked as A, the speed will has the exact same value V but the velocity will now point downward (just like the second ball).

So if you think about it, the first ball, from point A to the ground, will behave exactly like the second ball (same initial speed, same height).

That is why the speeds will be the same when they reach the ground.

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3 years ago
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3 years ago
A shopper pushes a grocery cart 20.0 m at constant speed on level ground, against a 35.0 N frictional force. He pushes in a dire
svp [43]

Answer:

(A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

Explanation:

Given that,

Distance =  20.0 m

Frictional force = 35.0 N

Angle = 25.0°

(A). We need to calculate the work done on the cart by the friction

Using formula of work done

W_{fr} = -F\cdot d

Where, F = force

d = distance

Put the value into the formula

W_{fr}=-35.0\times20

W_{fr}=−700\ J

(B). The work done by the gravity is perpendicular to the direction of the motion

We need to calculate the work done on the cart by the gravitational force

Using formula of work done

W=fd\cos\theta

Put the value into the formula

W=35.0\times20\cos90

W=0\ J

(C). We need to calculate the work done on the cart by the shopper

Using formula of work done

W_{sh}=W_{net}-W_{fr}

Put the value into the formula

W_{sh}=0-(-700)

W_{sh}=700\ J

(D). We need to calculate the force the shopper exerts

Using formula of force

F_{sh}=\dfrac{W_{fr}}{d\cos\theta}

Put the value into the formula

F_{sh}=\dfrac{700}{20\cos25}

F_{sh}=38.61\ N

(E). We need to calculate the total work done on the cart

Using formula of work done

W_{cart}=W_{fr}+W_{sh}

Put the value into the formula

W_{cart}=700-(-700)

W_{cart}=0\ J

Hence, (A). The work done on the cart by the friction is -700 J.

(B). The work done on the cart by the gravitational force is 0 J.

(C). The work done on the cart by the shopper is 700 J.

(D). The force the shopper exerts 38.61 N.

(E). The total work done on the cart is zero.

6 0
3 years ago
Pierre cycles at 4 miles per hour and travels a distance of 13 miles. How long
amm1812

Answer:

His journey took him 3 hours 15 minutes.

Explanation: 4 miles every hour. So 1 hr is equal to 4 miles, 2 hrs is equal to 8 miles, 3 hrs is equal to 12 miles. Now he just has 1 miles left, and since it takes him a hour to cycle 4 miles, 60 divided by 4 is 15. Therefore, 1 mile is equal to 15 minutes.

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4 years ago
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Fudgin [204]

Answer: all the energy is rushed in the wires due to the swich turning on when touching it may start a fire or electricute

Explanation:

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