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lesya692 [45]
3 years ago
6

if a tank filled with water contains a block and the height of the water above point A within the block is 0.6meter, what's the

pressure at point A?(a)5.88kPa(b)7.84kPa(c)3.85kPa(d)9.80kPa
Physics
2 answers:
alina1380 [7]3 years ago
7 0

your answer is 5.88kPa

i just took the test

Fittoniya [83]3 years ago
6 0
The pressure exerted by a fluid solely relies on the depth or height of the fluid, its density, and the gravitational constant. These three are related in the equation:

Pressure = density x g x height

In the problem, point A is within the block inside the tank. The water above the block is assumed to be 0.6 meters. This gives a point A pressure of:

P = 1000 kg/m^3 * 9.81 m/s^2 * 0.6 m = 5,886 Pa or 5.88KPa
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3 years ago
A box is placed on a conveyor belt that moves with a constant speed of 1.05 m/s. The coefficient of kinetic friction between the
sveticcg [70]

Answer:

The box stops in 0.139 seconds, after moving 7.29cm (0.0729m) backwards relative to the belt.

Explanation:

As the box is initially at rest relative to the earth, it is moving backwards with a speed of 1.05m/s relative to the belt. Then, the frictional force acts on the box to make it stop relative to the belt. So, we first have to write the equations of motion of the box in each axis:

x: f_k=ma\implies a=\frac{f_k}{m} \\\\y: N-mg=0\implies N=mg

Since the frictional force f_k is equal to f_k=\mu_k N=\mu_k mg, then we have that the acceleration is:

a=\frac{\mu_k mg}{m}=\mu_k g

Now, from the definition of acceleration we get:

a=\frac{v-v_0}{t}\implies t=\frac{v-v_0}{a}

And, as the final velocity is zero because the box gets to a stop, we have:

t=-\frac{v_0}{a}=-\frac{v_0}{\mu_k g}

(Don't worry about the negative sign. It will disappear because the initial velocity is also negative, since we take the box initially moving backwards)

Then, plugging in the given values, we calculate the time:

t=-\frac{(-1.05m/s)}{0.770(9.81m/s^{2})}=0.139s

In words, the time the box takes to stop sliding relative to the belt is 0.139s.

The displacement of the box in this time, is given by the kinematics formula:

v^{2}=v_0^{2}+2ax\implies x=-\frac{v_0^{2}}{2\mu_kg}

Finally, we calculate the displacement:

x=-\frac{(1.05m/s)^{2} }{2(0.770)(9.81m/s^2)}=-0.0729m=-7.29cm

This means that the box moves 7.29cm backwards relative to the belt.

4 0
2 years ago
A wagon has a mass of 100 grams and an
olga2289 [7]

Answer:

Net force on the wagon is 200 N

Explanation:

As we know by Newton's II law that net force on the system of mass is given as product of mass and acceleration

Here we know that

mass = 100 kg

a = 2 m/s/s

now we have

F = ma

F = (100)(2)

F = 200 N

3 0
3 years ago
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