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Fofino [41]
4 years ago
11

How can you use a graph to represent the motion of an object?

Physics
1 answer:
Pavlova-9 [17]4 years ago
3 0
X axis = time
y axis = displacement

(a brainliest would be appreciated)
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An object on Earth weighs 600 N and on Mars 100 N. If its mass on Earth is approximately 60 kg what is the mass of the object on
Deffense [45]

<u>Answer:</u>

I think its

<em>10 kg</em>

plzz plzz plzz. mark as brainliest

6 0
3 years ago
In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles
alexandr402 [8]

Answer:

Explanation:

We shall apply conservation of mechanical energy

kinetic energy of alpha particle is converted into electric potential energy.

1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³

= 484.4 x 10⁻¹⁶

=48.4  x 10⁻¹⁵ m

8 0
4 years ago
The attraction between earth and the moon is an example of ________ force. the attraction between earth and the moon is an examp
Korolek [52]

The answer is gravitational force. The gravitational force between the earth and the moon is the similar as between any other two masses in space.

Newton clarified that the force of attraction between two masses is the outcome of the weight of object one multiplied by the weight of objects two multiplied by the gravitational constant divided by the space between the two masses squared. 

4 0
4 years ago
Read 2 more answers
In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a
Novay_Z [31]

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

8 0
3 years ago
After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that
almond37 [142]

After three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

<h3>What is Half-Life?</h3>

Half-Life refers to the time it takes for half the amount of a substance to disappear or change.

The nucleus of the atoms of radioactive elements disintegrate to half their starting amounts after every Half-Life.

After three half-lives one-eight of the original atoms remain.

Therefore, after three half-lives have elapsed, the amount of an 8.0 g sample of a radionuclide that remains undecayed is 1.0 g.

Learn more about Half-Life at: brainly.com/question/26689704

#SPJ1

7 0
2 years ago
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