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3 years ago

State the Newton's first law of motion in words and in mathematical form.

Physics

1 answer:

Effectus [21]3 years ago

5 0

Newton's First Law

* Newton's First Law of Motion states that a body will remain at rest or will continue to move at a constant velocity, unless an external force is applied.

* This means that in order for the acceleration of a body to change, there must be a net force applied to the body. Put another way, if the forces on an object balance, there will be no acceleration (the object will continue at the same speed).So, if we are told that a body is not accelerating (i.e. if it is moving at a constant velocity), we know that the resultant (overall) force in any one direction will be zero.

* Its an unbalanced force,

8N acting towards the left, and 20N acting towards the right.

F = F2 - F1

= 20N - 8N

= 12N

F = ma

a = F/m

a = 12N/5kg

a = 2.4 m/s^2

Therefore, the accelaration of the object is 2.4 m/s^2

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3 years ago

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3 years ago

A quarterback throws a football 40 yards in 4 seconds.what is the average speed the football

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3 years ago

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A ball is thrown horizontally from the top of a 55 m building and lands 150 m from the base of the building. Ignore air resistan

PtichkaEL [24]

Answer:

a) t =3.349 s

b) V_x,i = 44.8 m/s

c) V_y,f = 32.85 m/s

d) V = 55.55 m/s

Explanation:

Given:

- Total throw in x direction x(f) = 150 m

- Total distance traveled down y(f) = 55 m

Find:

a) How long is the rock in the air in seconds.

b) What must have been the initial horizontal component of the velocity, in meters per second?

c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?

d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?

Solution:

- Use the second equation of motion in y direction:

y(f) = y(0) + V_y,i*t + 0.5*g*t^2

- V_y,i = 0 (horizontal throw)

55 = 0 + 0 + 0.5*(9.81)*t^2

t = sqrt ( 55 * 2 / 9.81 )

t =3.349 s

- Use the second equation of motion in x direction:

x(f) = x(0) + V_x,i*t

150 = 0 + V_x,i*3.349

V_x,i = 150 / 3.349 = 44.8 m/s

- Use the first equation of motion in y direction:

V_y,f = V_y,i + g*t

V_y,f = 0 + 9.81*3.349

V_y,f = 32.85 m/s

- The magnitude of velocity of ball when it hits the ground is:

V^2 = V_y,f^2 + V_x,i^2

V = sqrt (32.85^2 + 44.8^2)

V = 55.55 m/s

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3 years ago

A guitar string is 0.620m long, and oscillates at 234Hz. What is the velocity of the waves in the string? m/s

9966 [12]

Answer:

v = 72.54 m/s

Explanation:

We have,

Length of a guitar string is 0.62 m

Frequency of a guitar string is 234 Hz

For guitar string,

$L=2\lambda\\\\\lambda=\dfrac{L}{2}\\\\\lambda=\dfrac{0.62}{2}\\\\\lambda=0.31\ m$

The velocity of the wave in the string is given by :

$v=f\lambda\\\\v=234\times 0.31\\\\v=72.54\ m/s$

So, the velocity of the waves in the string is 72.54 m/s.

3 0

4 years ago