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Evgesh-ka [11]
3 years ago
11

In tests on earth a lunar surface exploration vehicle (mass = 5.57 × 103 kg) achieves a forward acceleration of 0.207 m/s2. to a

chieve this same acceleration on the moon, the vehicle's engines must produce a drive force of 1.60 × 103 n. what is the magnitude of the frictional force that acts on the vehicle on the moon?
Physics
1 answer:
spayn [35]3 years ago
5 0
The answer is 9000         or b                          
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CD bisects AB at point G. If AE = BE, which equation must be true?
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AG = BG equation must be true
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4 years ago
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A 12.0-g cd with a radius of 5.81 cm rotates with an angular speed of 33.2 rad/s. what is its kinetic energy?
schepotkina [342]
Use the formula 1/2*I*w^2 where w is the angular speed and I is the rotaional inirtia (for disk it is .5*mass*radius^2).
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3 years ago
parallel-plate capacitor is made of two square plates 25 cm on a side and 1.0 mmapart. The capacitor is connected to a 50.0-V ba
noname [10]

Answer:

6.9 x 10^-7 J  

3.5 x 10^-7 J

Explanation:

<u>Identify the unknown:  </u>

The energies stored in the capacitor before and after the plates are pulled farther apart  

<u>List the Knowns: </u>

Voltage of the battery: V = 50 V

Area of the plates: A = 0.25 x 0.25 = 0.0625 m^2

Original distance between the plates: d = 1 mm = 10^-3 m

New distance between the plates: d = 2 mm = 2 x 10^-3 m

Permittivity of free space: ∈o = 8.85 x 10^-12 C^2/Nm^2-

<u>Set Up the Problem:   </u>

Capacitance of a parallel-plate capacitor:  

C=∈o*A/d

Energy stored in a capacitor:  

U_c=(1/2)*V^2*C

<u>Solve the Problem:   </u>

<u>Before the plates are pulled farther apart:  </u>

C = 8.85 x 10^-12 x 0.0625/10^-3 = 5•53 x 10^-10 F  

U_c = (1/2) x (50)^2 x 5.53 x 10^-10= 6.9 x 10^-7 J  

<u>After the plates are pulled farther apart:  </u>

C = 8.85 x 10^-12 x 0.0625/2*10^-3 = 2•77 x 10^-10 F  

U_c = (1/2) x (50)^2 x 2•77 x 10^-10 = 3.5 x 10^-7 J  

The energy decrease because the capacitance decrease, so the stored charge decrease and transferred to the battery  

3 0
3 years ago
you stand on a straight desert road at night and observe a vehicle approaching. this vehicle is equipped with two small headligh
inysia [295]

Answer:

The distance that you marginally able to discern that there are two headlights rather than a single light source is 6.084 km

Explanation:

Given:

d = distance = 0.679 m

λ = wavelength of the light = 537 nm = 537x10⁻⁹m

dp = pupil diameter = 4.81 mm = 0.00481 m

Question: What distance, in kilometers, are you marginally able to discern that there are two headlights rather than a single light source, dx = ?

For the separation of the peak from the central maximum it is:

sin\theta =\frac{\lambda }{d_{p} } =\frac{537x10^{-9} }{0.00481} =1.116x10^{-4}

In this case, the two small sources of the headlights have the same angle as the images that form inside the eye

d_{x} =\frac{d}{sin\theta } =\frac{0.679}{1.116x10^{-4} } =6.084x10^{3} m=6.084km

7 0
3 years ago
White light is shining on a green grape. The grape __.
I am Lyosha [343]
White light is shining on a green grape. The grape would absorb all colors except green and will reflect the green color around and into our eyes. That's why the grape is green. Hope this helped!
5 0
3 years ago
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