What strongly affects the movement of the air masses along the earths surface

Determine the velocity that a car should have while traveling around a frictionless curve of radius 100m and that is banked 20 d

alex41 [277]

Answer:

$v=18.89\frac{m}{s}$

Explanation:

From the free-body diagram for the car, we have that the normal force has a vertical component and a horizontal component, and this component act as the centripetal force on the car:

$\sum F_x:Nsin(20^\circ)=ma_c(1)\\\sum F_y:Ncos(20^\circ)=mg(2)$

Solving N from (2) and replacing in (1):

$N=\frac{mg}{cos(20^\circ)}\\(\frac{mg}{cos(20^\circ)})sin(20^\circ)=ma_c\\gtan(20^\circ)=a_c$

The centripetal acceleration is given by:

$a_c=\frac{v^2}{r}$

Replacing and solving for v:

$gtan(20^\circ)=\frac{v^2}{r}\\v=\sqrt{grtan(20^\circ)}\\v=\sqrt{9.8\frac{m}{s^2}(100m)tan(20^\circ)}\\v=18.89\frac{m}{s}$

4 0

4 years ago

In which direction do hurricanes in the southern hemisphere rotate?

charle [14.2K]

They should rotate clockwise

5 0

3 years ago

A collector of rare items is negotiating with a museum. He wants to sell a spear he claims to have been

viva [34]

Answer:

I'm not sure if you maybe forgot to add in some information but I can give you the equation you will need to solve this..! (just without the bit of infomration you might need)

$2=4(\frac{1}{2})^{\frac{1100}{t}}$

In this case you'll be solving for t and I can help you with your first step, you'll divide 2 by 4, getting you to:

$\frac{1}{2} =\frac{1}{2}^{\frac{1100}{t}$

5 0

3 years ago

Read 2 more answers

What is the magnitude of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the

Zepler [3.9K]

Answer:

The magnitude of the magnetic field vector is 1.91T and is directed towards the east.

The steps to the solution can be found in the attachment below.

Explanation:

For the charge to remain in the the earth' gravitational field the magnetic force on the charge must be equal to the earth's gravitational force on the charge and must act opposite the direction of the earth's gravitational force.

Fm = Fg

qvBSin(theta) = mg

Where q = magnitude of charge

v = magnitude of the velocity vector = 4 x10^4 m/s

B = magnitude of the magnetic field vector

theta = the angle between the magnetic field and velocity vectors = 90°

m = mass of the charge = 0.195g

g = acceleration due to gravity =9.8m/s²

On substituting the respective values of all variables in the equation (1) above

B = 1.91T

The direction of the magnetic field vector was found by the application of the right hand rule: if the thumb is pointed in the direction of the magnetic force and the index finger is pointed in the direction of the velocity vector, the middle finger points in the direction of the magnetic field.

Below is the step by step procedure to the solution.