All categories

3 years ago

What force would be needed to make a 10kg bowling ball accelerate down the alleyway with an acceleration of 3m/s2

Physics

1 answer:

Anna007 [38]3 years ago

7 0

Answer:

30 N

Explanation:

From Newton's second law:

F = ma

F = (10 kg) (3 m/s²)

F = 30 N

You might be interested in

Fuel which has tremendous highest calorific value. (11 points if u answer this)

Vitek1552 [10]

Semi anthracite has the higest which is 29.5

8 0

3 years ago

Why a body in uniform velocity is zero acceleration ?

seropon [69]

Uniform velocity means no Net force and therefore no acceleration. Acceleration only happens when the velocity changes.

4 0

3 years ago

a coil has n turns enclosing an area of a. in a physics laboratory experiment, the coil is rotated during the time interval δt f

Mrac [35]

The correct answer for the question that is being presented above is this one:

Phi = BAsin(theta)
1. Phi(i) = BA 
2. Phi(f) = 0 
3. EMF = N(phi(i)-phi(f))/deltat

Here are the follow-up questions:

1. What is the total magnitude Phi_initial of the magnetic flux through the coil before it is rotated? 
2. What is the magnitude Phi_final of the total magnetic flux through the coil after it is rotated? 
3. What is the magnitude of the average emf induced in the coil?

8 0

3 years ago

Restate newtons first law of motion

andriy [413]

Newton's first law of motion states that an object that is resting (not moving) will stay motionless unless an external force acts on it and an object in motion will stay in motion unless an outside force acts on it. This law is known as newton's law of inertia.

7 0

3 years ago

A thin lens with a focal length of 6.0 cm is used as a simple magnifier by (a) what angular magnification is obtainable with the

Serhud [2]

Answer:

4.167

4.83871 cm

Explanation:

u = Object distance

v = Image distance = 25 cm

f = Focal length = 6 cm

Angular magnification is given by

$m=\frac{25}{f}\\\Rightarrow m=\frac{25}{6}\\\Rightarrow m=4.167$

The angular magnification of the lens is 4.167

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{6}=\frac{1}{u}+\frac{1}{-25}\\\Rightarrow \frac{1}{6}+\frac{1}{25}=\frac{1}{u}\\\Rightarrow \frac{1}{u}=\frac{31}{150}\\\Rightarrow u=4.83871\ cm$

The closest distance by which the object can be examined is 4.83871 cm

5 0

3 years ago