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pychu [463]
3 years ago
9

Prove that Efficiency=MA/ VR x 100%

Physics
1 answer:
geniusboy [140]3 years ago
5 0

I HOPE IT WILL HELP YOU. Have a great day ahead.

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An object with a mass of 70 kilograms is supported at a height 8 meters above the ground. What's the potential energy of the obj
Fofino [41]
The potential energy of an object is defined as the energy it harnesses as a result of the downwards pull of gravity. The acceleration due to gravity is equivalent to 9.8 m/s^2. Potential energy can be calculated using the formula: PE = mgz

where:
m = mass
g = acceleration due to gravity
z = height

Therefore, for this problem, the PE = 70(9.8)(8) = 5488 Joules
5 0
3 years ago
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A cylindrical shell of radius 7.00 cm and length 2.44 m has its charge uniformly distributed on its curved surface. The magnitud
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Answer:

(b) Use approximate relationships to find theelectric field at a point 4.00 cm from the axis, measured radiallyoutward from the midpoint of the shell.

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3 years ago
The car salesman tells you that the car can go from a stop position to 60 mph in six seconds is giving you the car’s rate of
slava [35]
The salesman is telling you the average magnitude of the car's acceleration.

| Acceleration | = (change in speed) / (time for the change)

| Acceleration | = (60 mi/hr) / (6 sec)

| Acceleration | =  10 miles/hr-sec

That would be 36,000 miles per hour squared,
or 0.0028 mile per second squared.
5 0
4 years ago
A proton moves along the x-axis in the laboratory with velocity uy = 0.6c. An observer moves with a velocity of v=0.8c along the
Dima020 [189]

Explanation:

Given that,

Velocity of the proton in lab frame u_{x} = 0.6c

Velocity of the observer v= 0.8c

We need to calculate the velocity of the proton with respect to the observer

Using formula of velocity

u'=\dfrac{v-u_{x}}{1-\dfrac{u_{x}v}{c^2}}

u'=\dfrac{0.8c-0.6c}{1-\dfrac{0.6c\times0.8c}{c^2}}

u'=c(\dfrac{0.8-0.6}{1-(0.8\times0.6)})

u'=0.385c

(a). We need to calculate the total energy of the proton in the lab frame

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u_{x}^2}{c^2}}}-1)

Where, Proton mass energy = m₀c²

Put the value into the formula

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.6c)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.6)^2}}-1)

K.E=234.57\ MeV

(b). We need to calculate the kinetic energy of the proton in the observer

Using formula of kinetic energy

K.E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{u'^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-\dfrac{(0.385)^2}{c^2}}}-1)

K.E=938.28\times(\dfrac{1}{\sqrt{1-(0.385)^2}}-1)

K.E=78.366\ MeV

(c). We need to calculate the momentum of the proton with respect to observer

Using formula of momentum

P_{obs}=\dfrac{m_{0}u'^2}{\sqrt{1-\dfrac{u'^2}{c^2}}}

We know that,

Proton mass energy = m₀c²

m_{0}=\dfrac{938.28}{c^2}

P_{obs}=\dfrac{\dfrac{938.28}{c^2}\times(0.385c)^2}{\sqrt{1-\dfrac{(0.385c)^2}{c^2}}}

P_{obs}=\dfrac{938.28\times(0.385)^2}{\sqrt{1-0.385^2}}

P_{obs}=150.69\ MeV/c

Hence, This is required solution.

3 0
3 years ago
Which of these is formed by a hot spot in Earth's crust?
dimulka [17.4K]

Answer:

we need to know what the choices are?

Explanation:

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3 years ago
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