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nevsk [136]
3 years ago
5

A satellite orbiting Mars (6.39x1023kg) experiences a gravitational field of 2.40N/kg. How far from Mars is the satellite orbiti

ng?
Physics
1 answer:
Ann [662]3 years ago
4 0

Answer:

sorry dont know

Explanation:

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Larry , Moe, and Curly are pushing on a 25 kg crate. The crate is sitting on a horizontal floor, and the coefficient of kinetic
monitta

Answer:

2 m/s².

Explanation:

The forces acting on the crate have been shown in the figure below.

Net force acting on the crate in the horizontal direction = 400 - 250 = 150 N towards right.

For the forces of vertical direction we can write,

Normal force = 150 + Mg = 150 + 250 = 400 N (here g = 10 m/s² have been taken)

The magnitude of frictional force that will act on the crate = μ × Normal force

So, Frictional force = 0.25 × 400 = 100 N

Thus, the net force that will cause acceleration in the crate will be

Force in the right direction - frictional force

= 150 - 100 = 50 N

As we know,

Acceleration = \frac{Force}{Mass} =  \frac{50}{25} = 2 m/s²

So, the net acceleration in the crate will be 2 m/s².

7 0
4 years ago
My first question friends please help me <br> Write the answer with steps !!!
telo118 [61]

Answer:

3.5mm I guess

Explanation:

More professional people will give you the correct answer... this is just a try

6 0
3 years ago
The image is the question
enyata [817]

Answer:

B

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3 0
3 years ago
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The benguela current, the peru current, and the california current is what
kvasek [131]
Part of the North Pacific Gyre
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4 years ago
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When ultraviolet light with a wavelength of 400 nm falls on a certain metal surface, the maximum kinetic energy of the emitted p
topjm [15]
In the photoelectric effect, the energy carried by the incoming photon is used partially to extract the photoelectron from the material  and the rest is converted into kinetic energy of the electron:
hf = \phi + K_{max} (1)
where
h is the Planck constant
f is the photon frequency
\phi is the work function of the material (the energy needed to extract the photoelectron)
K_{max} is the maximum kinetic energy of the photoelectron

The ultraviolet light has a wavelength of \lambda=400 nm=400 \cdot 10^{-9} m, so its frequency is
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{400 \cdot 10^{-9} m}=7.5 \cdot 10^{14}Hz
And the energy of each photon of this light is
E=hf=(6.6 \cdot 10^{-34}Js)(7.5 \cdot 10^{14}Hz)=4.95 \cdot 10^{-19}J

The maximum kinetic energy of the emitted photoelectrons is (converting into Joules)
K_{max} = 1.10 eV  \cdot 1.6 \cdot 10^{-19} J/eV =1.76 \cdot 10^{-19}J

And so by using equation (1) we can find the work function of the material:
\phi = hf - K_{max} = 4.95 \cdot 10^{-19}J - 1.76 \cdot 10^{-19} J =3.19 \cdot 10^{-19}J

Now we have instead light wavelength \lambda=270 nm= 270 \cdot 10^{-9}m hitting the same surface. The frequency of this light is
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{270 \cdot 10^{-9}m}=1.11 \cdot 10^{15} Hz
and the energy of each photon of this light is
E=hf=(6.6 \cdot 10^{-34}Js)(1.11 \cdot 10^{15}Hz)=7.33 \cdot 10^{-19} J

And so we can calculate the new maximum kinetic energy of the photoelectrons by using (1) and the work function we found previously:
K_{max} = hf-\phi = 7.33 \cdot 10^{-19}J - 3.19 \cdot 10^{-19}J = 4.14 \cdot 10^{-19}J

And if we want to convert it into electronvolts,
K_{max} =  \frac{4.14 \cdot 10^{-19}J}{1.6 \cdot 10^{-19} J/eV}=2.59 eV 

The maximum kinetic energy of the photoelectrons has increased because the light hitting the surface is now more energetic, so it can transfer more energy to the electrons.
8 0
3 years ago
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