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defon
3 years ago
6

Bunyi yang teratur disebut dengan

Physics
1 answer:
IgorLugansk [536]3 years ago
4 0
What language is this . i can translate , help
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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:

<em>Gravitational force on mass </em>m_{2}<em> due to mass </em>m_{1}<em> is given by </em>

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, g<em>ravitational force on mass </em>m_{2}<em> due to mass </em>m_{3}<em> is given by </em>

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
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