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3 years ago

14

Elias serves a volleyball at a velocity of 16 m/s. The mass of the volleyball is 0.27 kg. What is the height of the volleyball a

bove the gym floor if its total mechanical energy is 41.70 J? Round to the nearest tenth. m

1 answer:

liberstina [14]3 years ago

6 0

Answer:

The correct answer to the following question will be "2.7 m".

Explanation:

The given values are:

Velocity, v = 16 m/s

mass, m = 0.27 kg

Mechanical energy, M.E = 41.70 J

g = 9.8 m/s²

As we know,

Kinetic energy, $K=\frac{mv^2}{2}$

Potential energy, $U=m.g.h$

Now, the total mechanical energy will be:

⇒ $\frac{mv^2}{2}+U$

⇒ $41,71 \ kg.m^2/s^2$

Now,

⇒ $h=\frac{E-(\frac{mv^2}{2})}{mg}$

On putting the estimated values, we get

⇒ $=\frac{41.70-(\frac{0.27(16)^2}{2})}{0.27\times 9.8}$

⇒ $=\frac{41.70-(\frac{0.27\times 256}{2} )}{2.64}$

⇒ $=\frac{41.70-34.56}{2.64}$

⇒ $=\frac{7.14}{2.64}$

⇒ $=2.7 \ m$

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A small meteorite with mass of 1 g strikes the outer wall of a communication satellite with a speed of 2Okm/s (relative to the s

strojnjashka [21]

Answer:

The energy coverted to heat is 200 kilojoules.

Explanation:

GIven the absence of external forces exerted both on the small meteorite and on the communication satellite, the Principle of Linear Momentum is considered and let suppose that collision is completely inelastic and that satellite is initially at rest. Hence, the expression for the satellite-meteorite system:

$m_{M}\cdot v_{M} + m_{S}\cdot v_{S} = (m_{M}+m_{S})\cdot v$

Where:

$m_{M}$ , $m_{S}$ - Masses of the small meteorite and the communication satellite, measured in kilograms.

$v_{M}$ , $v_{S}$ - Speeds of the small meteorite and the communication satellite, measured in meters per second.

$v$ - Final speed of the satellite-meteorite system, measured in meters per second.

The final speed of the satellite-meteorite system is cleared:

$v = \frac{m_{M}\cdot v_{M}+m_{S}\cdot v_{S}}{m_{M}+m_{S}}$

If $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ and $v_{S} = 0\,\frac{m}{s}$ , the final speed is now calculated:

$v = \frac{(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)}{1\times 10^{-3}\,kg+200\,kg}$

$v = 0.1\,\frac{m}{s}$

Which means that the new system remains stationary and all mechanical energy from meteorite is dissipated in the form of heat. According to the Principle of Energy Conservation and the Work-Energy Theorem, the change in the kinetic energy is equal to the dissipated energy in the form of heat:

$K_{S} + K_{M} - K - Q_{disp} = 0$

$Q_{disp} = K_{S}+K_{M}-K$

Where:

$K_{S}$ , $K_{M}$ - Initial translational kinetic energies of the communication satellite and small meteorite, measured in joules.

$K$ - Kinetic energy of the satellite-meteorite system, measured in joules.

$Q_{disp}$ - Dissipated heat, measured in joules.

The previous expression is expanded by using the definition for the translational kinetic energy:

$Q_{disp} = \frac{1}{2}\cdot [m_{M}\cdot v_{M}^{2}+m_{S}\cdot v_{S}^{2}-(m_{M}+m_{S})\cdot v^{2}]$

Given that $m_{M} = 1\times 10^{-3}\,kg$ , $m_{S} = 200\,kg$ , $v_{M} = 20000\,\frac{m}{s}$ , $v_{S} = 0\,\frac{m}{s}$ and $v = 0.1\,\frac{m}{s}$ , the dissipated heat is:

$Q_{disp} = \frac{1}{2}\cdot \left[(1\times 10^{-3}\,kg)\cdot \left(20000\,\frac{m}{s} \right)^{2}+(200\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-(200.001\,kg)\cdot \left(0.001\,\frac{m}{s} \right)^{2}\right]$ $Q_{disp} = 200000\,J$

$Q_{disp} = 200\,kJ$

The energy coverted to heat is 200 kilojoules.

4 0

3 years ago

Two people must have the same speed and velocity if they are walking towards each other, they are walking away from each other,

iVinArrow [24]

Jogging side by side since the speed is equal and the direction is the same i.e same velocity

3 0

3 years ago

A quarter is tossed up from the roof of a skyscraper and hits the sidewalk below. Which of the following graphs best shows the v

laila [671]

Answer:

pic

Explanation:

5 0

2 years ago

If time travel to the future existed, doesn't that mean that the future has already occurred and that we are living in the past?

kodGreya [7K]

Answer:

As of right now the techology has not been invented to time travel

if we were to time travel to the future from where that person travled from would be the past and to them the people from where they came from are living in the past

Explanation:

3 0

3 years ago

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago