3 years ago

Help please Gravitation

Physics

1 answer:

xxMikexx [17]3 years ago

6 0

Answer:a force of attraction exerted by each particle of matter in the universe on every other particle

Explanation:gravitation describes the attractive force between any two masses

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This is the substance(s) formed in a chemical reaction.

vfiekz [6]

This is the substance(s) formed in a chemical reaction. is called the products

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3 years ago

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A rocket travels 1.3 km in 62 ms. What is its average speed in m⋅s−1? Do not give your answer in scientific notation. The answer

hjlf

Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

$Speed = \frac {distance}{time}$

$Speed = \frac {1300}{3720}$

Average speed = 0.35 m/s

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3 years ago

The driver of a car hears the siren of an ambulance that is moving away from her. If the actual frequency of the siren is 2000 h

MatroZZZ [7]

The answer is A.) 1,900 Hz

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3 years ago

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Calculate the pressure when a force of 200N acts on an area of 0.05cm2<br> .

deff fn [24]

FORCE ACTING ON THE BODY , F = 200N

AREA ON WHICH THE FORCE IS ACTING , A = 0.05 CM²

= 0.05*(10⁻²)

= 0.05*10⁻² M²

TO CONVERT CM TO M * BY 10⁻²

PRESSURE = FORCE / AREA

PRESSURE = 200/0.05*10⁻⁴

= 200*10⁴/0.05

= 200*10⁴*10²/5

= 20*10⁵/5

= 4*10⁵ N/M².

HOPE THIS HELPS.

4 0

2 years ago

Two tiny conducting spheres are identical and carry charges of -18.8 µC and +46.5 µC. They are separated by a distance of 2.47 c

sergiy2304 [10]

Answer:

$F=-12896N$ , attractive.

Explanation:

For calculating this force we use the Coulomb Law:

$F=\frac{kq_1q_1}{r^2}$

Where $k=9\times10^9Nm^2/C^{-2}$ is the Coulomb's constant, $q_1$ and $q_2$ the values of each charge and r the distance between them.

Since the Coulomb's constant as I wrote it is in S.I. we have to write all the magnitudes in that system of units, and substitute:

$F=\frac{(9\times10^9Nm^2/C^{-2})(-18.8\times10^{-6}C)(46.5\times10^{-6}C)}{(0.0247m)^2}=-12896N$

This force is attractive since both charges are of opposite sign.

3 0

3 years ago