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3 years ago

1

Physics

1 answer:

Darina [25.2K]3 years ago

3 0

Answer:

7.5 x 10⁻⁸N

Explanation:

Given parameters:

Mass 1 = 60kg

Mass 2 = 75kg

Distance between the bodies = 2m

Unknown:

Gravitational fore = ?

Solution:

The gravitational force between the two bodies can be derived using;

F = $\frac{G mass 1 x mass 2}{distance^{2} }$

G is the universal gravitation constant = 6.67 x 10⁻¹¹m³kg⁻¹s⁻²

Insert the parameters and solve;

F = $\frac{6.67 x 10^{-11} x 60 x 75}{2^{2} }$ = 7.5 x 10⁻⁸N

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3. Which of the following is the correctly abbreviated SI unit describing the mass of an object?

Paha777 [63]

Your answer will be A) KILOGRAMS

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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

Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a ligh

guapka [62]

Answer:

This is because the acceleration of objects due to gravity is independent of the mass of the object and is constant for all objects, therefore, all objects fall with the same speed.

Explanation:

The weight of an object or force of gravity acting on an object on the surface of earth is a product of its mass and acceleration due to gravity.

Mathematically, w = mg

where, m=mass of the object; g = acceleration due to gravity

Also, from newton's law of gravitation, gravitational force on the object ,F = GMm/r²

where G is the gravitational constant; M is mass of Earth; m is mass of object; r is the distance of separation between the object and the center of mass of the earth which is approximately the radius of earth.

Since the weight of an object is equal to the force of gravitation acting on it

W = F

mg = GMm/r²

g = GM/r²

The expression above is that of the relationship between the force of gravity acting on a body on the earth's surface, the weight of that body and the acceleration due to gravity, g.

It can be seen that the acceleration due to gravity g is independent of the mass of the object. Therefore, the acceleration of objects due to gravity is constant for all objects and all objects fall with the same speed.

4 0

3 years ago

Read 2 more answers

an 1150kg elevator moving down speeds up at a rate of 3.5m/s. what is the tension in the supporting cables?

gtnhenbr [62]

Answer:

The tension force in the supporting cables is 7245N

Explanation:

There are two forces acting on the elevator: the force of gravity pointing down (+) with magnitude (elevator mass) x (gravitational acceleration), and the tension force of the cable pointing up (-) with an unknown magnitude F. The net force is the sum of these forces:

$F_{net} = F_g - F = m\cdot g - F\\$

We are given the resulting acceleration along with the mass, i.e., we know the net force, allowing us to solve for F:

$1150kg\cdot 3.5\frac{m}{s^2}= 1150kg \cdot 9.8\frac{m}{s^2}-F\\\implies F = 1150kg\cdot(9.8-3.5)\frac{m}{s^2}= 7245N$

The tension force F in the supporting cables is 7245N

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4 years ago

Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s

Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density = 0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

= -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level ) = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = $\sqrt{CRT}$ = $\sqrt{1.4*287*200.5 }$ = 283.8 m/s

hence it is a subsonic aircraft

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3 years ago