What is the acceleration of the object?

A bucket filled woth water seems light while it sinks into water.Also show their relation using formula

slavikrds [6]

Answer: This phenomenon happens due to upthrust exerted by water.

Explanation:

We know that,

Liquid Pressure is directly proportional to the height of the vertical column in the liquid.(P∝h)

When a bucket filled water is sunk into the water container, there occur difference in the pressure in top and bottom of the water container. Due to this, water exerts an upward force on the bucket filled with water. This is called Uprthrust.

Upthrust on the bucket makes the bucket filled with water lose some of it's weight and causes apparent loss in weight.

Hence, the bucket filled with water seems light while it sinks into water.

6 0

3 years ago

What is the economical application of change of state of matter

kogti [31]

Answer:

Changes of state are physical changes in matter. They are reversible changes that do not change matter's chemical makeup or chemical properties. For example, when fog changes to water vapor, it is still water and can change back to liquid water again.

4 0

3 years ago

A 11.3-kg object oscillates at the end of a vertical spring that has a spring constant of 2.20 ✕ 104 N/m. The effect of air resi

3241004551 [841]

Answer:

(a) the frequency of the dampened oscillation is 7.02 Hz

(b) percentage decrease in amplitude of the oscillation in each cycle is 2%

Explanation:

Given;

mass of the object = 11.3 kg

the spring constant = 2.2 X 10⁴ N/m.

damping coefficient b = 3.00 N · s/m

Part (a) the frequency of the dampened oscillation

The oscillation frequency is calculated as follows;

$\omega _D = \sqrt{\omega_o^2 -(\frac{b}{2m})^2}\\\\\omega_o^2 = \frac{k}{m} =\frac{2.2X10^{4}}{11.3} = 1946.903rad/s\\\\thus, \omega _D = \sqrt{1946.903-(\frac{3}{2*11.3})^2} =44.12 rad/s$

The damped frequency $= \frac{\omega _D}{2\pi } = \frac{44.12}{2\pi } = 7.02 Hz$

Part (b) percentage decrease in amplitude of the oscillation in each cycle

The amplitude of the oscillation depends on the damping coefficient (b) and period (T), and it is given as;

$A(t) = e^{-\frac{b}{2m}(t)}$

After one cycle, the amplitude changes from A(t) to A(t+T), where T is the period of the oscillation.

$A(t +T) = e^{-\frac{b}{2m}(t+T)}$

Percentage decrease in amplitude is gotten by dividing A(t) by A(t+T)

$= \frac{e^{-\frac{b}{2m}(t)}}{e^{-\frac{b}{2m}(t+T)}} =e^{-\frac{b}{2m}(T)}$

But T = 1/f

Substituting the values of the parameters in the above equation, we will have;

$=e^{-\frac{b}{2m}(T)} = e^{-\frac{3}{2X11.3}(\frac{1}{7.02})} = 0.98$

Percentage decrease = 1 - 0.98 = 0.02 = 2%

4 0

4 years ago

A net torque applied to an object causes __________.

podryga [215]

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ A net toque applied to an object causes the angular velocity off the object to change.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

5 0

3 years ago

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10

Nuetrik [128]

Answer:

acceleraions 5.76g and 20.55g

Explanation:

This constant acceleration exercise can be solved using the kinematic equations in one dimension

Vf = Vo + a t

As part of the rest Vo = 0

a = Vf / t

a = 282/5

a = 56.4 m / s2

In relation to the acceleration of gravity

a ’= a / g = 56.4 / 9.8

a ’= 5.76g

To calculate the acceleration to stop we use the same formula

a2 = 282 / 1.40

a2 = 201.4 m / s2

This acceleration of gravity acceleration function is

a2 ’= 201.4 / 9.8

a2 ’= 20.55g

3 0

3 years ago