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stepan [7]
3 years ago
5

Using the right amount of significant figures, calculate the answer to the following problem, 215.5+101.02555

Physics
1 answer:
cestrela7 [59]3 years ago
7 0

Answer:

This is how I figured it out:

  1. 215.5 rounded to one significant figure is 200
  2. 101.02555 rounded to one significant figure is 100.
  3. 200 + 100 = 300.

Hope this helps!

Explanation:

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You push a refrigerator with a force of 100 N. If you move the refrigerator a distance of 5 m while you are pushing, how much wo
ddd [48]

The work done to push the refrigerator is 500 Nm.

Explanation:

Work done is the measure of force required to move any object from one point to another. So it is calculated as the product of force and displacement.

If the force increases the work done will increase and similarly, the increase in displacement increases the work done. So to push the refrigerator work should be done on the object and not by the object.

As the force is 100 N and the displacement is 5 m then, work done can be measured as

Work = Force × Displacement

Work = 100 × 5 = 500 Nm

So the work done to push the refrigerator is 500 Nm.

7 0
3 years ago
If the octupus had twelve arms then how many arms would the crab have show your work
Alexandra [31]

Answer:

I believe  that the crab would have 14 arms if the octopus had 12 arms

Explanation:

4 0
2 years ago
A raging bull of mass 700 kg runs at 36 km/h .How much kinetic energy does it have ?
Ipatiy [6.2K]

Answer:

Use equation for kinetic energy: Ek=mV²/2

m=700 kg

V=10m/s

Ek=700kg*100m²7s²/2

Ek=35000 J=35kJ

Explanation:

Hope this helps you

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5 0
3 years ago
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Which statement is the most appropriate conclusion for David to write?
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4 0
3 years ago
A charge of -3.02 μC is fixed in place. From a horizontal distance of 0.0377 m, a particle of mass 9.43 x 10^-3 kg and charge -9
Andreyy89

Answer:

d = 0.0306 m

Explanation:

Here we know that for the given system of charge we have no loss of energy as there is no friction force on it

So we will have

U + K = constant

\frac{kq_1q_2}{r_1} + \frac{1}{2}mv_1^2 = \frac{kq_1q_2}{r_2} + \frac{1}{2}mv_2^2

now we know when particle will reach the closest distance then due to electrostatic repulsion the speed will become zero.

So we have

\frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{0.0377} + \frac{1}{2}(9.43 \times 10^{-3})(80.4)^2 = \frac{(9 \times 10^9)(3.02 \mu C)(9.78 \mu C)}{r} + 0

7.05 + 30.5 = \frac{0.266}{r}

r = 7.08 \times 10^{-3} m

so distance moved by the particle is given as

d = r_1 - r_2

d = 0.0377 - 0.00708

d = 0.0306 m

6 0
3 years ago
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