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Alla [95]
4 years ago
5

Which part is not necessary to complete a circuit

Physics
2 answers:
Levart [38]4 years ago
6 0
A SWITCH I THINK??..

kondaur [170]4 years ago
6 0
A switch is the answer
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A 2.0-kilogram ball rolls down a ramp. if the ball accelerates at a rate of 12 m/s2, the net force causing the acceleration is:
ehidna [41]

Answer:

D

Explanation:

f = ma

2 x 12 = 24

answer could differ since it's rolling down a ramp. if an angle is given our approach differs.

7 0
2 years ago
Please help!
andrew-mc [135]

Answer:

V=W/Q

107V= W/17C

= We= 107×17 J

= 1819 J

Explanation:

hope it helps

3 0
3 years ago
In a bumper car arena, two cars of equal mass are heading straight toward each other. The orange one is traveling at a speed of
marshall27 [118]

Answer: D.

Explanation:   Orange, at 3 meters per second if you calculate the net force being applied to the system.

hope this helps! ✌

4 0
3 years ago
In your physics lab you are given a 10.1-kg uniform rectangular plate with edge lengths 68.7 cm by 47.5 cm. Your lab instructor
VladimirAG [237]

Answer:

2.35 kgm^2

Explanation:

we take length 68.7 cm as x-axis and 47.5 cm as y-axis then the axis about which we have to find out moment of inertia will be z-axis.

moment of inertia about x-axis

I_x = ML^2 /3 = 10.1\times 0.4752 /3 = 0.7596 kg-m2

I_y = 10.1\times 0.6872 / 3 = 1.5889 kgm^2

by perpendicular axis theorem

I_z = I_x + I_y = 0.7596 + 1.5889 = 2.35 kgm^2

4 0
3 years ago
The masses of the two moons are determined to be 2M2M for Moon AA and MM for Moon BB . It is observed that the distance between
seraphim [82]

Answer:

 F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

        F = G \frac{m M}{r^{2} }

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

        F_A = G \frac{2m M}{r^{2} }

Moon B

        F_B = G \frac{m M}{(2r)^{2} }

         F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

         F_B / F_A = \frac{1}{2 \ 4 } = 1/8

         F_A = 8 F_B

7 0
3 years ago
Read 2 more answers
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