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Scrat [10]
3 years ago
15

What is the correct formula for of the weight w of an object? assume that its mass is m, the magnitude of its acceleration is a,

and the local acceleration due to gravity is g?
Physics
2 answers:
Vsevolod [243]3 years ago
6 0
Weight = Mass × Acceleration
Step2247 [10]3 years ago
4 0
W = m • g

Weight = mass x local gravity.

The object's acceleration doesn't matter.
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During crystallisation the crystals separate out from the hot ________solution of a substance on cooling
alekssr [168]

Answer:

The process of separation or deposition of crystals from a hot saturated solution on gentle cooling of the solution is called 'crystallisation'.

Explanation:

6 0
3 years ago
Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's
mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

T^2\alpha R^3\\T^2=kR^3.......................(1)

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

\frac{T^2}{R^3}=k.......................(2)

Let the orbital period of the earth be T_e and its mean distance of from the sun be R_e.

Also let the orbital period of the planet be T_p and its mean distance from the sun be R_p.

Equation (2) therefore implies the following;

\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)

We make the period of the planet T_p the subject of formula as follows;

T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

R_p=16R_e...............(5)

Substituting equation (5) into (4), we obtain the following;

T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}

R_e^3 cancels out and we are left with the following;

T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)

Recall that the orbital period of the earth is about 365.25 days, hence;

T_p=64*365.25\\T_p=23376days

4 0
3 years ago
30 POINTS!
brilliants [131]

Answer:

d

Explanation:

5 0
3 years ago
Read 2 more answers
The __________-second rule applies to any speed in ideal weather and road conditions.
SOVA2 [1]
The two-second rule.

It is a common guideline to follow while driving.

It means that any given driver should be AT LEAST two seconds behind any vehicle that is driving in front of his vehicle. It might apply for any kind of vehicle.
8 0
3 years ago
A goal should NOT be:
Korvikt [17]

Answer: unspecific

Explanation:

you should always be specific with your goals, say someones goal is to make it to state for swimming, he/she should be specific to the event in which they want to go to state in.

6 0
3 years ago
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