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vodomira [7]
3 years ago
6

Which two formulas are used to calculate potential and kinetic energy

Physics
2 answers:
goblinko [34]3 years ago
6 0

Answer:

gravitational potential energy:

GPE = m g h

kinetic energy:

KE = 1/2 m v^2

kogti [31]3 years ago
3 0

Answer:

\boxed{\bold { \large { \boxed {KE=\frac{1}{2} mv^2 \ , \ PE=mgh}}}}

Explanation:

Kinetic energy formula

\displaystyle KE=\frac{1}{2} mv^2

Potential energy formula

\displaystyle PE=mgh

\displaystyle KE \Rightarrow \sf kinetic \ energy \ (J)

\displaystyle PE \Rightarrow \sf potential \ energy \ (J)

\displaystyle m \Rightarrow \sf mass \ (kg)

\displaystyle v \Rightarrow \sf velocity \ (m/s)

\displaystyle g \Rightarrow \sf acceleration \ of \ gravity\ (m/s^2)

\displaystyle h \Rightarrow \sf height \ (m)

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Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed
Ivanshal [37]

Answer:

Part a)

f_w = f_g = 4.57 \times 10^{14} Hz

Part b)

\lambda_w = 492 nm

\lambda_g = 437.3 nm

Part c)

v_w = 2.25 \times 10^8 m/s

v_g = 2.0 \times 10^8 m/s

Explanation:

Part a)

frequency of light will not change with change in medium but it will depend on the source only

so here frequency of light will remain same in both water and glass and it will be same as that in air

f = \frac{v}{\lambda}

f = \frac{3 \times 10^8}{656 \times 10^{-9}}

f = 4.57 \times 10^{14} Hz

Part b)

As we know that the refractive index of water is given as

\mu_w = 4/3

so the wavelength in the water medium is given as

\lambda_w = \frac{\lambda}{\mu_w}

\lambda_w = \frac{656 nm}{4/3}

\lambda_w = 492 nm

Similarly the refractive index of glass is given as

\mu_w = 3/2

so the wavelength in the glass medium is given as

\lambda_g = \frac{\lambda}{\mu_g}

\lambda_g = \frac{656 nm}{3/2}

\lambda_g = 437.3 nm

Part c)

Speed of the wave in water is given as

v_w = \frac{c}{\mu_w}

v_w = \frac{3 \times 10^8}{4/3}

v_w = 2.25 \times 10^8 m/s

Speed of the wave in glass is given as

v_g = \frac{c}{\mu_g}

v_g = \frac{3 \times 10^8}{3/2}

v_g = 2 \times 10^8 m/s

4 0
3 years ago
Starting from rest, a person runs with a constant acceleration, traveling 40 meters in 10 seconds. What is their final velocity?
Assoli18 [71]

Answer:

Final velocity v = 8.944 m/sec

Explanation:

We have given distance S = 40 meters

Time t = 10 sec

As it starts from rest so initial velocity u = 0

From second equation of motion s=ut+\frac{1}{2}at^2

40=0\times 10+\frac{1}{2}a10^2

a=0.8944m/sec^2

Now from first equation of motion v=u+at, here v is final velocity, u is initial velocity, a is acceleration and t is time

So v=u+at=0+0.8944\times 10=8.944m/sec

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zepelin [54]

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