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Allushta [10]
2 years ago
8

If 5000Pa a pressure is exerted on an object with the contact area 0.04m^2. Calculate the mass of an object

Physics
1 answer:
Whitepunk [10]2 years ago
6 0
  • Pressure = 5000 Pa
  • Contact Area = 0.04 m^2
  • Acceleration due to gravity = 9.8 m/s^2
  • Let the force be F.
  • We know, Force = Pressure × Contact Area
  • Therefore, Force = 5000 Pa × 0.04 m/s^2
  • or, Force = 200 N
  • We know, force = mass × acceleration
  • Therefore, mass = force ÷ acceleration
  • or, mass = 200 N ÷ 9.8 m/s^2 = 20.4 Kg

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>.</u><u>4</u><u> </u><u>Kg</u>

Hope you could understand.

If you have any query, feel free to ask.

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Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

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3 years ago
A ball is kicked at an angle of 35° with the ground.a) What should be the initial velocity of the ball so that it hits a target
stiks02 [169]

Answer:

a.18.5 m/s

b.1.98 s

Explanation:

We are given that

\theta=35^{\circ}

a.Let v_0 be the initial velocity of the ball.

Distance,x=30 m

Height,h=1.8 m

v_x=v_0cos\theta=v_0cos35

v_y=v_0sin\theta=v_0sin35

x=v_0cos\theta\times t=v_0cos35\times t

t=\frac{30}{v_0cos35}

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Substitute the values

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1.8=30tan35-\frac{6574.6}{v^2_0}

\frac{6574.6}{v^2_0}=21-1.8=19.2

v^2_0=\frac{6574.6}{19.2}

v_0=\sqrt{\frac{6574.6}{19.2}}=18.5 m/s

Initial velocity of the ball=18.5 m/s

b.Substitute the value then we get

t=\frac{30}{18.5cos35}

t=1.98 s

Hence, the time for the ball to reach the target=1.98 s

7 0
3 years ago
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