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dimulka [17.4K]
3 years ago
11

A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W

ater flows out from the bottom through a small hole. How fast is this water moving?
Physics
1 answer:
weqwewe [10]3 years ago
7 0

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

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Blake and Sandra are having a rummage sale. Blake drags 3 boxes a distance of 10 meters each. He exerts a force of 20 newtons on
vampirchik [111]
For Blake:

3 boxes at a distance of 10 meters each, each box weighs 20 N

Work done by Blake = 3 * 10m * 20N
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For Sandra:

4 boxes, 15 N each at a distance of 12 meters each. 

Work done by Sandra = 4 * 15 N *12m
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Power = 720 J/ 4 min
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Blake does less work than Sandra. 
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5 0
3 years ago
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A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the
Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

W_{r} = |F_{r}|*|d|*cos(\theta_{1})

Where:                

F_{r}: is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

W_{g} = |F_{g}|*|d|*cos(\theta_{2})  

The force of gravity is given by the weight of the bike.

F_{g} = -mgsin(24)     

And the angle between the force of gravity and the direction of motion is 180°.

W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})  

W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J  

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.  

I hope it helps you!                                                                                          

3 0
3 years ago
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Nata [24]
Gain energy

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A person's _____________ will change if they move from the Earth to the moon.
Diano4ka-milaya [45]

1. A person's weight will change if they move from the Earth to the moon.

In fact, the weight of a person is given by:

W=mg

where m is the mass of the person and g is the gravitational acceleration. The mass of the person, m, is the same on the Earth and on the moon, but the value of g is different on the Moon (about 1/6 of the Earth's value), so the weight also changes.


2. An astronaut is launched into space. The mass of the astronaut did not change. This is measured in Kg.

The mass of an object (or of a person, as in this case) is an intrinsec property of the object, that depends on the amount of matter inside the object: therefore, this quantity does not depend on the location of the object, so it is the same on the Earth, on the Moon and in space.


3. What is the weight of a ring tailed lemur that has a mass of 10 kg? -98 N

The weight of the lemur is given by:

W=mg

where m=10 kg is the lemur's mass and g=-9.8 m/s^2 is the gravitational acceleration. Using these numbers, we find

W=(10 kg)(-9.8 m/s^2)=-98 N

and the negative sign simply means that the direction of the weight is downward.


4. What is the mass of the lemur from the previous question if it was on the International Space Station? 10 kg

As we said in question 1), the mass of an object does not depend on the location, so the mass of the lemur is still 10 kg, as in the previous exercise.


5. A rocket being thrust upward as the force of the fuel being burned pushes downward is an example of which of Newton's laws? Third's Newton Law

Third's Newton Law states that:

"When an object A exerts a force on an object B, then object B exerts an equal and opposite force on object A".

Applied to this case, the two objects are the fuel and the rocket. The fuel is pushed backward by the rocket, so the fuel exerts an equal and opposite force on the rocket, which then moves forward.


6. When a cannon is fired, the projectile moves forward. According to Newton's 3rd law, the cannon will want to travel backward.

Third's Newton Law states that:

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Applied to this case, the two objects are the cannon and the projectile.The projectile is pushed forward by the cannon, so the projectile exerts an equal and opposite force on the cannon, which moves backward.


7. An object has a weight of 21,532 N on Earth. What is the mass of the object? 2,197 kg

The weight of the object is given by: W=mg

If we re-arrange the formula and we use W=21,532 N, we can find the mass of the object:

m=\frac{W}{g}=\frac{21,532 N}{9.8 m/s^2}=2,197 kg


8. What is the mass of the object from the previous question if we put it on the moon? The force of gravity on the moon is 1.62 m/s2.  2,197 kg

As we said in question 4), the mass of an object does not change if we move it to another location, so its mass is still 2,197 kg.


9. How much force is exerted if a 250 kg object has an acceleration of 750 m/s2 ? 187,500 N

The force exerted on the object is given by Newton's second law:

F=ma

where F is the force, m=250 kg is the mass and a=750 m/s^2 is the acceleration. By using these numbers, we find

F=(250 kg)(750 m/s^2)=187,500 N


10. A resting soccer ball moving after it is kicked is an example of which of Newton's laws? Newton's second law

Newton's second law states that when an object is acted upon unbalanced force, the object has an acceleration, given by the law

F=ma

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5 0
3 years ago
QUESTION 3
Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately <u>2.937 m/s²</u>.

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, \mu_s = 0.5

Wood on wood kinetic friction, \mu_k = 0.3

Solution;

The force of friction, F_f, acting when the crate is moving is given as

follows;

F_f = m × g × \mu_k

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

F_f = 100 × 9.81 × 0.3 = 294.3

The force of friction, F_f = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

a = \dfrac{F}{m}

Therefore;

a = \dfrac{293.7 \ N}{100 \ kg} = 2.937

The acceleration of the crate, a ≈ <u>2.937 m/s²</u>.

Learn more about friction here:

brainly.com/question/94428

8 0
2 years ago
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