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GrogVix [38]
3 years ago
13

2. Can you place three forces of 5g, 6g, and 12g so they are in equilibrium. Justify your answer.

Physics
1 answer:
Bond [772]3 years ago
4 0

Answer:

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

Explanation:

Equilibrium means their sum must be zero.

Here the forces are 5g, 6g, and 12g.

For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.

Here

        Magnitude of largest force = 12 g

        Sum of magnitudes of other forces = 5g + 6g = 11g

       Magnitude of largest force >   Sum of magnitudes of other forces

So this forces cannot form equilibrium.

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

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Scattering occurs when light changes direction after colliding with particles of matter.
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The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter
GenaCL600 [577]

Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

Q= required charge

Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

diameter of the conductor is 4.20 mm

But Radius= diameter/2= 4.20/2=

The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then

radius of the conductor is 0.0021m.

We can now calculate the area of the conductor which is

A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

We can proceed to calculate the current density below

J = 140/13.85*10^-6 = 10108303A/m

According to the listed reference:

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s

Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

7 0
3 years ago
What is a change in the speed of an object called?
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Acceleration........................................

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Can classical physics be used to accurately describe a satellite moving at a speed of 7500 m/s? explain why or why not.
gogolik [260]

The classical physics works on the Newton's laws of motion. It is applicable on heavenly bodies which are governed by the gravitational force. On the other hand, Quantum Physics is applicable for very low mass and sized bodies like electron, protons etc. The classical physics would accurately describe the motion of satellite moving with speed 7500 m/s using the following formula:

r=\frac {GM}{v^{2}}

where G is the gravitational constant, M is the mass of the planet and v is the orbital speed. Then radius of the orbit can be described by this formula.

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3 years ago
What is the final velocity of the ball that is dropped from a height of 200m?
Hoochie [10]

The final velocity of the ball that is dropped from a height of 200m is v = 44.73 m/s .

<h3>What is velocity with example?</h3>

The rate at which an object is travelling in one direction is referred to as its velocity. an automobile traveling north on a highway, or a rocket taking off. Its velocity vector's absolute value always is equal to the motion's speed because it is a scalar.

<h3>Briefing:</h3>

Given the initial velocity of the ball (u) =  0

Distance travelled by the ball (s) = 200m

Acceleration (a) = 10 m/s²

As we know:

v² = u² + 2as

Putting values:

v² = 0+2 × (10 m/s²) × (200 m)

v = 44.73 m/s.

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