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maxonik [38]
2 years ago
13

A ball rolls 60 centimeters along a sidewalk in 5 seconds what is the speed of the ball.

Physics
2 answers:
Reika [66]2 years ago
8 0

Answer:

12 cm/second

Explanation:

To calculate the speed of an object, simply obtain the distance traveled by the ball and divide it by the total time it took to travel 60 centimeters.

⇒ Distance/Time = Speed

In this case, the ball traveled 60 centimeters in 5 seconds.

<u>Therefore, the speed of the ball is;</u>

60 centimeters (Distance)/5 seconds (Time) = 12 cm/second

meriva2 years ago
6 0

Answer:

12 cm/s

Explanation:

Quite simply, you are looking for  cm/s

  so   60 cm / 5s   = 12 cm/s

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1)The position change of almost any manually operated room light switch.

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Explanation:

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3 years ago
A person who normally weighs 580 N is riding in an elevator that is moving upward, but slowing down at a steady rate. If this pe
Zanzabum

Answer:

M au = Fs - M g       au = upwards acceleration; Fs = scale reading

Fs = M (au + g)    scalar quantities where g is positive downwards and au is positive upwards - Fs is the net force acting on the person

If the acceleration is zero Fs = M g  and the scale reads the persons weight

If the elevator is decelerating then au is negative and the scale reading     Fs = (g - au) M     and the scale reading is less than the weight of the person

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2 years ago
A vector is 14.4 m long and
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Answer:

Explanation:

The x-component is found in the magnitude of the vector times the cosine of the angle.

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3 0
3 years ago
Protons and neutrons grouped in a specific pattern
alexgriva [62]
Answer b protons and electrons
5 0
3 years ago
What are (a) the charge and (b) the charge density on the surface of a conducting sphere of radius 0.20 m whose potential is 240
Lady bird [3.3K]

Answer:

(a) charge q=5.33 nC

(b) charge density σ=10.62 nC/m²

Explanation:

Given data

radius r=0.20 m

potential V=240 V

coulombs constant k=9×10⁹Nm²/C²

To find

(a) charge q

(b) charge density σ

Solution

For (a) charge q

As

V_{potential}=kq/r\\ q=rV_{potential}/k\\q=\frac{(0.20)(240)}{9*10^{9} }\\ q=5.333*10^{-9}C\\or\\ q=5.33nC

For (b) charge density

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σ=10.62 nC/m²

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