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Ber [7]
3 years ago
6

An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

nd hits the ground 1.37 s later. Part A) What is the acceleration due to gravity on this asteroid?
Physics
1 answer:
SCORPION-xisa [38]3 years ago
5 0

Answer:

a_y=0.92m/s^2

Explanation:

To solve this problem we use the formula for accelerated motion:

y=y_0+v_{y0}t+\frac{a_yt^2}{2}

We will take the initial position as our reference (y_0=0m) and the downward direction as positive. Since the rock departs from rest we have:

y=\frac{a_yt^2}{2}

Which means our acceleration would be:

a_y=\frac{2y}{t^2}

Using our values:

a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2

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Suppose a light source is emitting red light at a wavelength of 700 nm and another light source is emitting ultraviolet light at
klasskru [66]

Answer:

b) twice the energy of each photon of the red light.

Explanation:

\lambda = Wavelength

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

Energy of a photon is given by

E=h\nu\\\Rightarrow E=h\dfrac{c}{\lambda}

Let \lambda_1 = 700 nm

\lambda_2=350\\\Rightarrow \lambda_2=\dfrac{\lambda_1}{2}

For red light

E_1=\dfrac{hc}{\lambda_1}

For UV light

E_2=\dfrac{hc}{\dfrac{\lambda_1}{2}}

Dividing the equations

\dfrac{E_1}{E_2}=\dfrac{\dfrac{hc}{\lambda_1}}{\dfrac{hc}{\dfrac{\lambda_1}{2}}}\\\Rightarrow \dfrac{E_1}{E_2}=\dfrac{1}{2}\\\Rightarrow E_2=2E_1

Hence, the answer is  b) twice the energy of each photon of the red light.

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3 years ago
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Contrasting the Doppler effect with shock waves, the one that requires the faster source is?
MAXImum [283]

I think shock waves require more speed they travel at the speed of sound

6 0
3 years ago
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A skateboarder with a mass of 60 kg moves with a force of 20 N. What is her acceleration?
Zanzabum

Explanation:

Solution,

  • Mass(m)= 60 kg
  • Force (F)= 20 N
  • Acceleration (a)= ?

We know that,

  • F=ma
  • a=F/m
  • a=20/60
  • a=0.333 m/s²

So, her acceleration is 0.333 m/s².

4 0
2 years ago
An electric field of magnitude 2.35 V/m is oriented at an angle of 25.0° with respect to the positive z-direction. Determine the
zzz [600]

Answer:

The magnitude of the electric flux is 3.53\ N-m^2/C

Explanation:

Given that,

Electric field = 2.35 V/m

Angle = 25.0°

Area A= 1.65 m^2

We need to calculate the flux

Using formula of the magnetic flux

\phi=E\cdot A

\phi = EA\cos\theta

Where,

A = area

E = electric field

Put the value into the formula

\phi=2.35\times1.65\times\cos 25^{\circ}

\phi=2.35\times1.65\times0.91

\phi=3.53\ N-m^2/C

Hence, The magnitude of the electric flux is 3.53\ N-m^2/C

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3 years ago
Energy that is passed on from one trophic level to the next is called what?
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Energy decreases as it moves uptrophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level.Trophic level transfer efficiency (TLTE) measures the amount of energy that is transferred between trophic levels.
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