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TEA [102]
3 years ago
9

Calculate the pressure increase at the bottom of a pool of depth 2m. Density of water=1000kg/m^3 g=10N/kg

Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
Please check the attached picture

Blababa [14]3 years ago
4 0
Pa = atmospheric pressure (look it up in units of N/m^2)
D = density of water (given as 1000 kg/m^3)
h = depth of water = 3m
g = acceleration of gravity = 9.8 m/s^2

At a depth of h (meters), the static pressure is

P(h) = Pa + Dgh

Plug in the numbers and get P(h) in units of N/m^2.
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g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart
Veseljchak [2.6K]

Answer:

The speed of q₂ is 4\sqrt{10}\ m/s

Explanation:

Given that,

Distance = 0.4 m apart

Suppose, A small metal sphere, carrying a net charge q₁ = −2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q₂ = −8μC and mass 1.50g, is projected toward q₁. When the two spheres are 0.800m apart, q₂ is moving toward q₁ with speed 20m/s.

We need to calculate the speed of q₂

Using conservation of energy

E_{i}=E_{f}

\dfrac{1}{2}mv_{i}^2+\dfrac{kq_{1}q_{2}}{r_{i}}=\dfrac{kq_{1}q_{2}}{r_{f}}+\dfrac{1}{2}mv_{f}^2

\dfrac{1}{2}m(v_{i}^2-v_{f}^2)=kq_{1}q_{2}(\dfrac{1}{r_{f}}-\dfrac{1}{r_{i}})

Put the value into the formula

\dfrac{1}{2}\times1.5\times10^{-3}(20^2-v_{f}^2)=9\times10^{9}\times-2\times10^{-6}\times-8\times10^{-6}(\dfrac{1}{(0.4)}-\dfrac{1}{(0.8)})

0.00075(400-v_{f}^2)=0.18


400-v_{f}^2=\dfrac{0.18}{0.00075}

-v_{f}^2=240-400

v_{f}^2=160

v_{f}=4\sqrt{10}\ m/s

Hence, The speed of q₂ is 4\sqrt{10}\ m/s

7 0
3 years ago
Noooooooooooooooooooooooo
Afina-wow [57]

Answer:

yes

Explanation:

5 0
3 years ago
Read 2 more answers
In the formula vy = ayt, what is the value of ay for an object in projectile motion?
Sergeu [11.5K]
–9.8 m/s<span>2

Just took it and got it right!</span>
8 0
3 years ago
Read 2 more answers
A rock is rolling down a hill. At position 1, its velocity is 2.0 m/s. Twelve seconds later, as it passes position 2, its veloci
alukav5142 [94]
As this happens over twelve seconds, you would take the total difference in velocities and divide it by twelve to find the change per second

44.0 m/s - 2.0 m/s = 42.0 m/s 

42.0 m/s / 12 s = 3.5 m/s2

the acceleration of the rock would be 3.5 m/s2
4 0
3 years ago
Read 2 more answers
Please help! This is due tomorrow and I absolutely need help.
zysi [14]

Answer:

Correct answer:  11. Total distance d = 200m ; 12. Vav = 3.63m/s ;

13. Total displacement Dt = 0m ; 14. V₂(10s-15s) = 0 m/s ;

15. V₃(15s-40s) = 4 m/s ; 16. V₁(0s-10s) = 6 m/s > V₄(40s-55s) = 2.67 m/s

Explanation:

The whole movement can be divided into four stages.

In the first stage the subject moves 60m in a positive direction for 10s,

in the other it is stationary for 5s, in the third it moves 100m in the opposite (negative) direction for 25s and in the fourth in the positive 40m for 15s.

11. Total distance = 60 + 0 + 100 + 40 = 200m

12. The formula for calculating the average speed (velocity) is

Vav = (S₁ + S₂ + S₃ + S₄) / (t₁ + t₂ + t₃ + t₄)

Vav = (60 + 0 + 100 + 40)/ (10 + 5 + 25 + 15) = 200/55 = 3.63 m/s

13. The movement started from the origin and ended at the origin

Total displacement is zero meters.

14. The speed between 10s and 15s is zero, because he did not move.

15. V₃ = S₃/t₃ = 100/25 = 4 m/s

16. V₁ = S₁/t₁ = 60/10 = 6 m/s   and V₄ = S₄/t₄ = 40/15 = 2.67 m/s

V₁ > V₄

God is with you!!!

4 0
3 years ago
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