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Kamila [148]
3 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]3 years ago
7 0

Answer:

300000

Explanation:

p=30x10x1000=30000pascal

kaheart [24]3 years ago
4 0

Answer:

300000

p=30x10x1000=30000pascal

Explanation:

U WILL TRHANK ME LATER!

MARK ME AS BRAINLISAT!

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anyanavicka [17]
Because if your putting tension on something tensions obviously going to increase with more pressure and weight on it
7 0
3 years ago
HELP PLS I WILL MARK U BRAINLIEST!!
Alex17521 [72]

Answer:

D

Explanation:

I would say much too small as there is a significant portion of the plant that is UNDER water.....

    (  Really depends on how deep the water is...if it is shallow water it would be just a little too small)

8 0
2 years ago
Please help!! giving a lot of points
Readme [11.4K]

Question 1.

  • mass = 4500 kg
  • potential energy (p.e) = 67500 J

now, we know :

=》

p.e =  mgh

=》

67500 = 4500 \times 10 \times h

=》

67500 = 45000 \times h

=》

h =  \dfrac{67500}{45000}

=》

h = 1.5 \: m

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

  • mass = 4500 kg
  • kinetic energy = 63000 j

we know,

=》

k.e =  \dfrac{1}{2} mv {}^{2}

=》

63000 =  \dfrac{1}{2}  \times 4500 \times  {v}^{2}

=》

{v}^{2}  =  \dfrac{63000 \times 2}{4500}

=》

{v}^{2}  = 28

=》

v =  \sqrt{28}

=》

v = 2 \sqrt{7} \:  \:  ms {}^{ - 1}

or

=》

5.29 \:  \: ms {}^{ - 1}

7 0
3 years ago
You apply a net force on a soccer ball of 15 N. If the acceleration it has is 5 m/s2 what is the mass of the ball?​
Ipatiy [6.2K]

Answer:

<h2>3 kg </h2>

Explanation:

The mass of the ball can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

We have

m =  \frac{15}{5}  = 3 \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

4 0
3 years ago
A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg
anyanavicka [17]

Answer:

0.4113772 s

Explanation:

Given the following :

Mass of bullet (m1) = 8g = 0.008kg

Initial horizontal Velocity (u1) = 280m/s

Mass of block (m2) = 0.992kg

Maxumum distance (x) = 15cm = 0.15m

Recall;

Period (T) = 2π√(m/k)

According to the law of conservation of momentum : (inelastic Collison)

m1 * u1 = (m1 + m2) * v

Where v is the final Velocity of the colliding bodies

0.008 * 280 = (0.008 + 0.992) * v

2.24 = 1 * v

v = 2.24m/s

K. E = P. E

K. E = 0.5mv^2

P.E = 0.5kx^2

0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2

0.5*1*5.0176 = 0.5*k*0.0225

2.5088 = 0.01125k

k = 2.5088 / 0.01125

k = 223.00444 N/m

Therefore,

Period (T) = 2π√(m/k)

T = 2π√(0.992+0.008) / 233.0444

T = 2π√0.0042910

T = 2π * 0.0655059

T = 0.4113772 s

6 0
3 years ago
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