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2 years ago

A bowling ball is dropped from a height of 24 feet.

Physics

1 answer:

sdas [7]2 years ago

7 0

Answer:

a. H=(1/2)*a*t^2

B.h=16.09ft

C.t= 1.22s

Explanation:

Explanation: Newton's equation of motion. This is called motion under free fall.

S=ut +(1/2)*a*t^2

Let S=H, and a=g

H=ut +(1/2)*g*t^2

H=distance covered in ft(24ft)

u= initial velocity (ft/s)

t= time covered in seconds

g=acceleration due to gravity(ft/s^2)

a=g (32.185ft/s)

When u=0

H=(1/2)*g*t^2

2. H=(1/2)*g*t^2

g=32.185ft/s^2

t=1s

Juxtaposing the values into the equation

H= (1/2)* 32.185*(1)^2

H=16.09ft

3. When the maximum distance covered is 24ft, then using

H=(1/2)*a*t^2

24=0.5*32.165*t^2

24=16.09t^2

1.491=t^2

Find the square root of both sides

t=1.22s

It takes 1.22s to hit the ground

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Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of ea

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Explanation:

Work done by gravity is given by the formula,

W = $\rho A (h_{1} - h)g (h - h_{2})$ ......... (1)

It is known that when levels are same then height of the liquid is as follows.

h = $\frac{h_{1} + h_{2}}{2}$ ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

W = $\frac{1}{4} \rho gA(h_{1} - h_{2})^{2})$

= $\frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})$

= 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

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3 years ago

The liquid mercury in a thermometer expands as it is heated. this is a...

dedylja [7]

Change of State and a Physical Change

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2 years ago

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Discuss how a photon (aka the light particle) can be affected by gravity despite being massless.

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Explanation:

Light is clearly affected by gravity, just think about a black hole, but light supposedly has no mass and gravity only affects objects with mass. On the other hand, if light does have mass then doesn't mass become infinitely larger the closer to the speed of light an object travels.

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2 years ago

A builder drops a brick from a height of 15 m above the ground. The gravitational field strength g is 10 N/ kg. What is the spee

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The speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Given the data in the question;

Since the brick was initially at rest before it was dropped,

Initial Velocity; $u = 0$

Height from which it has dropped; $h = 15m$

Gravitational field strength; $g = 10N/kg = 10 \frac{kg.m/s^2}{kg} = 10m/s^2$

Final speed of brick as it hits the ground; $v = \ ?$

<h3>Velocity</h3>

velocity is simply the same as the speed at which a particle or object moves. It is the rate of change of position of an object or particle with respect to time. As expressed in the Third Equation of Motion:

$v^2 = u^2 + 2gh$

Where v is final velocity, u is initial velocity, h is its height or distance from ground and g is gravitational field strength.

To determine the speed of the brick as it hits the ground, we substitute our giving values into the expression above.

$v^2 = u^2 + 2gh\\\\v^2 = 0 + ( 2\ *\ 10m/s^2\ *\ 15m)\\\\v^2 = 300m^2/s^2\\\\v = \sqrt{300m^2/s^2}\\ \\v = 17.32m/s$

Therefore, the speed of the brick dropped by the builder as it hits the ground is 17.32m/s.

Learn more about equations of motion: brainly.com/question/18486505

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1 year ago

Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc

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Answer:

Temperature at the exit = $267.3 C$

Explanation:

For the steady energy flow through a control volume, the power output is given as

$W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

Inlet area of the turbine = $60cm^{2}= 0.006m^{2}$

To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume $v_{1}$

$v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg$

$m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec$

for Ideal gasses, the enthalpy change can be calculated using the formula

$h_{2}-h_{1}=C_{p}(T_{2}-T_{1})$

hence we have

$W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})$

$250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})$

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have $T_{2}=267.3C$

Hence, the temperature at the exit = $267.3 C$

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2 years ago