if a net horizontal force of 175 N is applied to a bike whos mass is 43 kg what acceleration is produced

An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

4 0

2 years ago

Which of the following best describes an action-reaction pair? A. The Moon Pulls on Earth, and Earth pulls back on the moon. B.

Papessa [141]

An action-reaction pair would be a pair in which one of the elements exerts a force on the other element (action), and then the other element would respond to this force by exerting another force in the opposite direction (reaction).

From the given choices, we will see that:
For choice A, the moon exerts a force on the earth by pulling it (action) and the earth responds to this force by pulling the moon (reaction in opposite direction of the action).

Therefore, the correct choice would be:
A. <span>The Moon Pulls on Earth, and Earth pulls back on the moon.</span>

4 0

2 years ago

Read 2 more answers

A cylinder of radius R and height H is floating upright in

emmainna [20.7K]

Answer:

Pressure difference between Top and Bottom of the cylinder is given as

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

Explanation:

As we know that the force due to pressure is balanced by the weight of the cylinder

So we will have

$F = mg$

so we have

$\Delta P \pi R^2 = mg$

so we have

$\Delta P \pi R^2 = \pi R^2(\rho_A(\frac{H}{2}) + \rho_B(\frac{H}{2}))g$

so we have

$\Delta P = \frac{gH}{2}(\rho_A + \rho_B)$

6 0

2 years ago

A camera gives a proper exposure when set to a shutter speed of 1/250 s at f-number F8.0. The photographer wants to change the s

Oksana_A [137]

Answer:

F4.0

Explanation:

To obtain a shutter speed of 1/1000 s to avoid any blur motion the f-number should be changed to F4.0 because the light intensity goes up by a factor of 2 when the f-number is decreased by the square root of 2.

5 0

2 years ago

A lens of focal length 15.0 cm is held 10.0 cm from a page (the object ). Find the magnification .

nevsk [136]

Answer:

Magnification, m = 3

Explanation:

It is given that,

Focal length of the lens, f = 15 cm

Object distance, u = -10 cm

Lens formula :

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

v is image distance

$\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-10)}\\\\v=-30\ cm$

Magnification,

$m=\dfrac{v}{u}\\\\m=\dfrac{-30}{10}\\\\m=3$

So, the magnification of the lens is 3.

3 0

2 years ago