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Bond [772]
2 years ago
10

Is any force exerted on the vertical sides of the loop that you used in the experiment and how does it affect the apparent mass?

Physics
1 answer:
lorasvet [3.4K]2 years ago
5 0
2. How should employers respond to K to 12 graduates who apply for vacant positions in
3. What were the perceived disadvantages of K to 12 graduates pcompared to college students?
4. What factors could give K to 12 graduates an advantage in the labor market?
Discussion Questions
1.
What is the dilemma K to 12 graduates face when applying for a job?
their company? Pa help asap po
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Neon, helium, methane, krypton, xenon, hydrogen, and ozone make up & what of the atmosphere
Alinara [238K]
Part of the upper atmosphere

6 0
3 years ago
A powerful missile reaches a speed of 5 kilometers per second in 10 seconds after its launch. What is the average acceleration o
AveGali [126]

Average acceleration  =  (change in speed)  /  (time for the change)

The missile's change in speed is  (5,000 - 0)  =  5,000 m/s

Average acceleration  =  (5,000 m/s)  /  (10 sec)

                                 =      500 m/s²       (about  51 Gs)

Inconveniently, this isn't one of the choices on the list. Is there something wrong, either with the choices or with my solution ?

No. Relax. Everything is OK.

500 meters is the same thing as 0.5 kilometer.  So my answer can also be written as  0.5 km/s² .  That doesn't change anything, and it IS one of the listed choices.

The average acceleration is <em>0.5 km/sec²  (d)</em> .

4 0
2 years ago
Read 2 more answers
A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.
Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

  1. Kinetic energy.
  2. Potential energy.

Kinetic energy =\frac{1}{2} mv^2

Potential energy =\frac{Kq_1q_2}{d}

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

                      =\frac{1}{2} mv^2+ \frac{Kq_1q_2}{d}

                     =\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}

                    =(1280-337.5)J

                    =942.5 J

Total energy of a system remains constant.

Therefore,

E =\frac{1}{2} mv^2 + \frac{Kq_1q_2}{d}

\Rightarrow  942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}

\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750

\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750

\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}

\Rightarrow v= 1961.19   m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

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How can you inhabit saturn
vaieri [72.5K]
I know
the answer do you want me to help you?
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While flying over the grand canyon the pilot
Veronika [31]
We cannot answer the question if we do not have the answers - sorry!!
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3 years ago
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