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Alex17521 [72]
3 years ago
13

How many significant figures are in the measurement 0.020 km?

Physics
2 answers:
nirvana33 [79]3 years ago
7 0

Answer:

2

Explanation:

0.020 has two significant figures.

The 2 is significant because all non-zero numbers are signficant.

The second 0 is significant because all zeros at the end of a decimal are significant.

The other zeros are not significant, because they are just place holders.

Volgvan3 years ago
4 0

Answer:3

Explanation:

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explain why balancing the torques acting on a body is not enough to establish equilibrium. give an example to justify your answe
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To establish equilibrium all torques need to be justified , the body needs to be in translational as well as rotational equilibrium.

<h3>What is equilibrium and why balancing the torques is not enough to establish equilibrium?</h3>
  1. Equilibrium is a condition of balance between forces, or influences from opposite ways .
  2. For a body to attain equilibrium , need to have both translational as well as rotational equilibrium.
  3. Force is responsible for translational equilibrium and torque is responsible for rotational equilibrium.
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1 year ago
Why is it that an object can accelerate while
posledela

For the same reason that you can skate around a curve at constant speed but not with constant velocity.

The DIRECTION you're going is part of your velocity, but it's not part of your speed.

If the DIRECTION changes, that's a change of velocity.

The object doesn't have to change speed to have a different velocity. A change of direction is enough to do it.

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3 years ago
A proton moves through an electric potential created by a number of source charges. Its speed is 2.5 x 105 m/s at a point where
Kaylis [27]

Answer:

Proton’s speed, a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

Explanation:

Given;

initial speed of proton, u = 2.5 x 10⁵ m/s

initial potential, V = 1500 V

mass of proton = 1.67 x 10⁻²⁷ kg

Work done, W = eV= ΔK.E = ¹/₂mu²

eV = ¹/₂mu² (J)

where;

e is the charge of the proton in coulombs

V is the electric potential in volts

m is the mass of the proton in kg

u is the speed of the proton in m/s

m =\frac{2eV_1}{u_1{^2}} = \frac{2eV_2}{u_2{^2}} = \frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}}

\frac{V_1}{u_1{^2}}} =\frac{V_2}{u_2{^2}} = \frac{1500}{(2.5*10^5)^2} = \frac{500}{u_2{^2}} \\\\u_2{^2} =\frac{500*(2.5*10^5)^2}{1500} = 0.333*6.25*10^{10}\\\\u_2 = \sqrt{0.333*6.25*10^{10}} =1.4 *10^5 \ m/s

Therefore, proton’s speed a short time later when it reaches a point of lower potential is 1.4 x 10⁵ m/s

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Hope this helps.
(Please mark this brainliest, I would really appreciate it) Thanks!
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