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3 years ago

11

A container is filled with water and the pressure at the container bottom is P. If the container is instead filled with a liquid

having specific gravity 1.05, what new bottom pressure will be measured

1 answer:

den301095 [7]3 years ago

7 0

Eat ahah is when s susuehs

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A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago

What does the term calorie refer to

OLga [1]

Hey there,
<span>It means- the amount of energy required to raise the temperature of a liter of water one degree centigrade at sea level
</span>
Hope this helps :))

~Top

6 0

4 years ago

When you drop a 0.43 kg apple, Earth exerts

Archy [21]

Answer:

The acceleration of the earth is 7.05 * 10^-25 m/s²

Explanation:

<u>Step 1:</u> Data given

mass of the apple = 0.43 kg

acceleration = 9.8 m/s²

mass of earth = 5.98 * 10 ^24 kg

<u>Step 2:</u> Calculate the acceleration of the earth

Following the third law of Newton F = m*a

F(apple) = F(earth) = m(apple)*a(apple)

F(apple) = 0.43 kg * 9.8 m/s² = 4.214 N

a(earth) = F(apple/earth)/m(earth)

a(earth) = 4.214N /5.98 * 10 ^24 kg

a(earth) = 7.05 * 10^-25 m/s²

The acceleration of the earth is 7.05 * 10^-25 m/s²

4 0

3 years ago

The inductance in the drawing has a value of L = 9.4 mH. What is the resonant frequency f0 of this circuit?

yuradex [85]

Answer:

The resonant frequency of this circuit is 1190.91 Hz.

Explanation:

Given that,

Inductance, $L=9.4\ mH=9.4\times 10^{-3}\ H$

Resistance, R = 150 ohms

Capacitance, $C=1.9\ \mu F=1.9\times 10^{-6}\ C$

At resonance, the capacitive reactance is equal to the inductive reactance such that,

$X_C=X_L$

$2\pi f_o L=\dfrac{1}{2\pi f_oC}$

f is the resonant frequency of this circuit

$f_o=\dfrac{1}{2\pi \sqrt{LC}}$

$f_o=\dfrac{1}{2\pi \sqrt{9.4\times 10^{-3}\times 1.9\times 10^{-6}}}$

$f_o=1190.91\ Hz$

So, the resonant frequency of this circuit is 1190.91 Hz. Hence, this is the required solution.

4 0

3 years ago

Describe four consequences of having a different type of joint in the thumb.

Mariulka [41]

Answer:

Essentially, your thumb is the main piece of your body that has saddle joints. The bones in your seats joint are in charge of moving forward and backward, side to side.

When all is said in done, the piece of the thumb joint that is subjected to extreme anxiety is that known as CMC joint or carpometacarpal joint. This joint is fundamentally shaped by the metacarpal bone and it explains with the trapezium bone of the wrist.

Explanation:

5 0

3 years ago