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Kamila [148]
2 years ago
5

The valve in the exit pipe is closed . The density of water is 1000kg / m and the gravitational force on free fall of water is 1

0N / Kg . Calculate the pressure of the water acting on the closed valve in the exit pipe 30m​
Physics
2 answers:
Stels [109]2 years ago
7 0

Answer:

300000

Explanation:

p=30x10x1000=30000pascal

kaheart [24]2 years ago
4 0

Answer:

300000

p=30x10x1000=30000pascal

Explanation:

U WILL TRHANK ME LATER!

MARK ME AS BRAINLISAT!

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Explanation:

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2 years ago
a car accelerates from 2 m/s to 28m/s at a constant rate of 3 m/s^2. How far does it travel while accelerating?
xxTIMURxx [149]

Answer:

Distance, S = 130m

Explanation:

Given the following data;

Initial velocity = 2m/s

Final velocity = 28m/s

Acceleration = 3m/s²

To find the distance, we would use the third equation of motion.

V² = U² + 2aS

Substituting into the equation, we have;

28² = 2² + 2*3*S

784 = 4 + 6S

6S = 784 - 4

6S = 780

S = 780/6

Distance, S = 130m

3 0
2 years ago
What must be the distance in meters between point charge q1 = 30.3 µC and point charge q2 = -68.2 µC for the electrostatic force
Zolol [24]

Answer: 1.62 m

Explanation: In order to solve this problem we have to use the definition of Coulomb force whic is given by:

F=(k*q1*q2)/d^2  where k is a constant 9*10^9 N/C^2*m^2. d is the distance between tha charges q1 and q2.

then we can:

d^2= (k*q1*q2)/F= (9*10^9*30.3*10^-6*68.2*10^-6)/7.09=1.62 m

(note we have  considered q2 as positive because we are determining d^2 by using the magnitude of the electric Force bewteen the charges).

4 0
2 years ago
Two bumper cars move in a straight line with the following equations of motion: x1 = -4.0 m + (1.1 m/s )t x2 = 8.8 m + (-2.9 m/s
lawyer [7]

Answer:

3.2 seconds

Explanation:

When the cars collide, they have the same position.

x₁ = x₂

-4 + 1.1t = 8.8 − 2.9t

4t = 12.8

t = 3.2

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2 years ago
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Svetlanka [38]
A colloid I think. Don’t hold it against me if I’m wrong my dude.
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