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Flura [38]
3 years ago
13

The liquid pressure varies with depth why​

Physics
1 answer:
Solnce55 [7]3 years ago
3 0

Explanation:

Hey, there!

The liquid pressure varies with depth because liquid pressure is directly proportional to the depth of liquid from the free surface of the liquid. so, more the depth more the pressure and less the depth less the pressure.

<em><u>Hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

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A ball is thrown straight up. What are the velocity and acceleration of the ball at the highest point in its path?
zubka84 [21]

Answer:

b. v = 0, a = 9.8 m/s² down.

Explanation:

Hi there!

The acceleration of gravity is always directed to the ground (down) and, near the surface of the earth, has a constant value of 9.8 m/s². Since the answer "b" is the only option with an acceleration of 9.8 m/s² directed downwards, that would solve the exercise. But why is the velocity zero at the highest point?

Let´s take a look at the height function:

h(t) = h0 + v0 · t + 1/2 g · t²

Where

h0 = initial height

v0 = initial velocity

t = time

g = acceleration due to gravity

Notice that the function is a negative parabola if we consider downward as negative (in that case "g" would be negative). Then, the function has a maximum (the highest point) at the vertex of the parabola. At the maximum point, the slope of the tangent line to the function is zero, because the tangent line is horizontal at a maximum point. The slope of the tangent line to the function is the rate of change of height with respect to time, i.e, the velocity. Then, the velocity is zero at the maximum height.

Another way to see it (without calculus):

When the ball is going up, the velocity vector points up and the velocity is positive. After reaching the maximum height, the velocity vector points down and is negative (the ball starts to fall). At the maximum height, the velocity vector changed its direction from positive to negative, then at that point, the velocity vector has to be zero.

8 0
3 years ago
Which best explains why a wood-burning fireplace represents an open system?
Alona [7]
The answer will be B
5 0
3 years ago
The graph above shows the position x as a function of time for the center of mass of a system of particles of total mass 6. 0 kg
kumpel [21]

The resulting change in momentum of the system will be +18.6 Ns. The momentum is conserved.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

The given data in the problem is;

m is the mass =6.0 kg

t is the time interval=2 second

From Newton's second law;

\rm \triangle P =m \triangle V \\\\ \triangle P= m(\frac{\triangle x}{\triangle t} )\\\\

From the graph;

\rm \triangle t = 2sec\\\\ \triangle x = (12-8) m

The change in the momentum is;

\rm \triangle P = m\tr(\frac{v-u}{t}) \\\\ \triangle P =9.3 \times \frac{12-8}{2} \\\\ \triangle P= +18.6 \  N.s

Hence, the resulting change in momentum of the system will be +18.6 Ns.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ1

6 0
1 year ago
how long will it take you to travel from rustburg hign school to watch the Devils beat the brookville bees if you live 20 miles
tresset_1 [31]

(20 miles) x ( 1/45  hour/mile) = 

       (20/45) (hour) = <em>

         4/9  hour =    </em>
26 minutes  40 seconds


7 0
2 years ago
What is the efficiency (w/qh) of an ideal carnot heat engine operating between a hot region at t= 400 k and a cold one at t= 300
Vinvika [58]
The efficiency of an ideal Carnot heat engine can be written as:
\eta = 1-  \frac{T_{cold}}{T_{hot}}
where
T_{cold} is the temperature of the cold region
T_{hot} is the temperature of the hot region

For the engine in our problem, we have T_{cold}=300 K and T_{hot}=400 K, so the efficiency is
\eta= 1 - \frac{300 K}{400 K}=0.25
4 0
2 years ago
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