Bunyi yang teratur disebut dengan

Three small spheres, having masses m1 = 1 kg, m2 = 3 kg, and m3 = 4 kg, are held fixed on the x axis in deep space where the eff

Vlad [161]

Answer: $F_{net} = 0.7411 N$ towards the mass $m_{3}$

Explanation: We know that the gravitational force is a long range force which is always attractive in nature.

Given that:

mass $m_{1} = 1 kg$

mass $m_{2} = 3kg$

mass $m_{3} = 4kg$

The masses are positioned on X-axis at the following points:

Position of mass $m_{1}$ $x_{1} = 0$

Position of mass $m_{2}$ $x_{2} = 3$

Position of mass $m_{3}$ $x_{3} = 6$

Mathematically:

Gravitational force on mass $m_{2}$ due to mass $m_{1}$ is given by

$F_{21} = G \frac{m_{1}.m_{2}}{(r_{21})^2}$ ...................(1)

where: $(r_{21})^2$ = the radial distance between masses $m_{2}$ & $m_{1}$ =3

Similarly, gravitational force on mass $m_{2}$ due to mass $m_{3}$ is given by

$F_{23} = G \frac{m_{3}.m_{2}}{(r_{23})^2}$ ............................(2)

where: $(r_{23})^2$ = the radial distance between masses $m_{2}$ & $m_{3}$ =3

Now, put the respective values in the above equations.

$F_{21} = 6.67 \times 10^{-11 }\times \frac{1\times 3}{3^2}$

$F_{21} = 2.2233\times 10^{-11} N$

Again,

$F_{23} = 6.67 \times 10^{-11 }\times \frac{1\times 4}{3^2}$

$F_{23} = 2.9644\times 10^{-11} N$

∵Mass $m_{2}$ is in the middle of the masses $m_{3}$ & $m_{1}$ therefore the forces $F_{23}$ & $F_{21}$ will attract them in radially opposite direction.

∴ $F_{net} = F_{23} -F_{21} \\\\F_{net} = 2.9644-2.2233\\\\F_{net} = 0.7411 N$ towards the mass $m_{3}$

6 0

3 years ago

Read 2 more answers

What is the term for the ability to cause change in matter?

Leokris [45]

Answer:

A. Energy

Explanation:

You can change energy from one form to another when you lift your arm or take a step. Energy is used to move matter. In this scenario, the matter would be you.

3 0

3 years ago

AS AIRPLANE WENT FROM 120 m/s to 180 m/s in 4.0 seconds what is the acceleration

marusya05 [52]

Answer:15metre per second squ

I

Explanation:

acceleration=(final velocity-initial velocity)÷t

acceleration=(180-120)÷4

acceleration=60÷4

acceleration=15 metre per second square

8 0

3 years ago

A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the gro

grin007 [14]

Answer:

torque is 1.7 * $10^{-2}$ Nm

Explanation:

Given data

turns n = 1000 turns

radius r = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

torque is 1.7 * $10^{-2}$ Nm

5 0

3 years ago

A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius o

Setler79 [48]

Answer:

21.8 m/s

Explanation:

At the top of the hill (crest), there are two forces acting on the motorcycle:

- The reaction force of the road, N (upward)

- The force of gravity, mg (downward)

Since the motorcycle is moving by circular motion, the resultant of these forces will give the centripetal force, so:

$mg-N = m\frac{v^2}{r}$

where the direction of the weight (mg) is equal to that of the centripetal force, and where

m is the mass of the cycle

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed

r = 48.6 is the radius of the hill

The cycle loses contact with the road when the reaction force becomes zero:

N = 0

Substituting into the equation, we therefore find the maximum speed that is allowed for the cycle before losing constact:

$mg = m\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8)(48.6)}=21.8 m/s$

6 0

3 years ago