All categories

2 years ago

A continental polar air mass forms in

Physics

1 answer:

Sedaia [141]2 years ago

4 0

Answer:

B. Northern Canada

Explanation:

A continental polar air mass can form over the land during the winter months. In the Northern Hemisphere, it originates in northern Canada or Alaska. As it moves southward, it brings dry weather conditions to the United States. Temperature and humidity levels are both low. Hope this helps :)

You might be interested in

A penny is dropped from the 82nd Floor Observatory of the Empire State Building. The penny will fall for 8.1s before hitting the

Alexxandr [17]

Answer:

vf=79.38 m/s

y= 321.5 m

Explanation:

Free Fall Motion

A free-falling object falls under the exclusive influence of gravity. Free-falling objects do not encounter air resistance.

If an object is dropped from rest in a free-falling motion, it falls with a constant acceleration called the acceleration of gravity, which value is $g = 9.8 m/s^2$ .

The final velocity of a free-falling object after a time t is given by:

vf=g.t

The distance traveled by a dropped object is:

$\displaystyle y=\frac{gt^2}{2}$

The penny will fall for t=8.1 s before hitting the ground, thus the height from which it was dropped is:

$\displaystyle y=\frac{9.8\cdot 8.1^2}{2}$

y= 321.5 m

The final velocity is:

$vf=9.8\cdot 8.1$

vf=79.38 m/s

8 0

2 years ago

A tennis player tosses a tennis ball straight up and then catches it after 1.77 s at the same height as the point of release. (a

Bogdan [553]

(a) 9.8 m/s^2, downward

There is only one force acting on the ball while it is in flight: the force of gravity, which is

F = mg

where

m is the mass of the ball

g is the gravitational acceleration

According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so

F = mg = ma

which means

a = g

So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:

g = -9.8 m/s^2

where the negative sign means the direction is downward.

(b) v = 0

Any object thrown upward reaches its maximum height when its velocity is zero:

v = 0

In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.

The initial velocity of the ball can be found by using the equation:

v = u + at

Where

v = 0 is the velocity at the maximum height

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration

t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so

$t=\frac{1.77 s}{2}=0.89 s$

So we can now solve the equation for u, and we find:

$u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s$

(d) 3.88 m

The maximum height reached by the ball can be found by using the equation:

$v^2 - u^2 = 2ad$

where

v = 0 is the velocity at the maximum height

u = 8.72 m/s is the initial velocity

a = g = -9.8 m/s^2 is the gravitational acceleration

d is the maximum height reached

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(8.72 m/s)^2}{2(-9.8 m/s^2)}=3.88 m$

7 0

2 years ago

A 2120 kg car traveling at 13.4 m/s collides with a 2810 kg car that is initally at rest at a stoplight. The cars stick together

viktelen [127]

Answer:

The coefficient of friction between the cars and the road is 0.859.

Explanation:

The two cars collide each other inelastically, then we can determine the resulting velocity by the Principle of Momentum Conservation:

$m_{A}\cdot v_{A} + m_{B}\cdot v_{B} = (m_{A} + m_{B})\cdot v$ (1)

Where:

$m_{A}$ , $m_{B}$ - Masses of the cars, in kilograms.

$v_{A}$ , $v_{B}$ - Initial velocities of the cars, in meters per second.

$v$ - Velocity of the resulting system, in meters per second.

If we know that $m_{A} = 2120\,kg$ , $v_{A} = 13.4\,\frac{m}{s }$ , $m_{B} = 2810\,kg$ and $v_{B} = 0\,\frac{m}{s}$ , then the velocity of the resulting system:

$v = \frac{m_{A}\cdot v_{A}+m_{B}\cdot v_{B}}{m_{A}+m_{B}}$

$v = \frac{(2120\,kg)\cdot \left(13.4\,\frac{m}{s} \right)+(2810\,kg)\cdot \left(0\,\frac{m}{s} \right)}{2120\,kg + 2810\,kg}$

$v = 5.762\,\frac{m}{s}$

By Principle of Energy Conservation and Work-Energy Theorem, we understand that the initial translational kinetic energy ( $K$ ), in joules, is dissipated due to work done by friction ( $W_{f}$ ), in joules, that is to say: