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3 years ago

How do altitude temperature and air pressure relate

2 answers:

7 0

They relate because the further up you go, the colder it gets, and the air pressure decreases the further up you go. The altitude temperature and the air pressure both decrease, and that is their relationship. Altitude temperature decreases, the higher you go, and air pressure also decreases, the higher up you go. Therefore, the lower down you go, the higher the air pressure, and the higher the altitude temperature.

Hope this helps and have a nice day:)

liberstina [14]3 years ago

6 0

The higher up you are, the lower those two factors are.

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A 300 kg car initially travels with a velocity of 20 m/s to the right. A net force F acts on the car for 5 s, which causes the v

hram777 [196]

Answer:

600N.

Explanation:

From the question, we are to calculate the net force acting on the car.

According to Newton's second law of motion:

F = ma

m is the mass of the car

a is the acceleration = change in velocity/Time

a = v-u/t

F = m(v-u)/t

v is the final velocity = 30m/s

u is the initial velocity = 20m/s

t is the time = 5secs

m = 300kg

Get the net force:

Recall that: F = m(v-u)/t

F = 300(30-20)/5

F = 60(30-20)

F = 60(10)

F = 600N

Hence the net force acting on the car is 600N.

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3 years ago

What is the main function of a telescope?

Korvikt [17]

It magnifies light received from distant objects.

4 0

3 years ago

Read 2 more answers

If the Sun were the size of a small exercise ball (about 0.5 meters (m) in

suter [353]

Answer:

the size of a pea

Explanation:

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3 years ago

Sort the forces as producing a torque of positive, negative, or zero magnitude about the rotational axis identified in part

Fantom [35]

a) Angular acceleration: $17.0 rad/s^2$

b) Weight: conterclockwise torque, reaction force: zero torque

Explanation:

In this problem, you are holding the pencil at its end: this means that the pencil will rotate about this point.

The only force producing a torque on the pencil is the weight of the pencil, of magnitude

$W=mg$

where m is the mass of the pencil and g the acceleration of gravity.

However, when the pencil is rotating around its end, only the component of the weight tangential to its circular trajectory will cause an angular acceleration. This component of the weight is:

$W_p =mg sin \theta$

where $\theta$ is the angle of the rod with respect to the vertical.

The weight act at the center of mass of the pencil, which is located at the middle of the pencil. So the torque produced is

$\tau = W_p \frac{L}{2}=mg\frac{L}{2} cos \theta$

where L is the length of the pencil.

The relationship between torque and angular acceleration $\alpha$ is

$\tau = I \alpha$ (1)

where

$I=\frac{1}{3}mL^2$

is the moment of inertia of the pencil with respect to its end.

Substituting into (1) and solving for $\alpha$ , we find:

$\alpha = \frac{\tau}{I}=\frac{mg\frac{L}{2}sin \theta}{\frac{1}{3}mL^2}=\frac{3 g sin \theta}{2L}$

And assuming that the length of the pencil is L = 15 cm = 0.15 m, the angular acceleration when $\theta=10^{\circ}$ is

$\alpha = \frac{3(9.8)(sin 10^{\circ})}{2(0.15)}=17.0 rad/s^2$

There are only two forces acting on the pencil here:

- The weight of the pencil, of magnitude $mg$

- The normal reaction of the hand on the pencil, R

The torque exerted by each force is given by

$\tau = Fd$

where F is the magnitude of the force and d the distance between the force and the pivot point.

For the weight, we saw in part a) that the torque is

$\tau =mg\frac{L}{2} cos \theta$

For the reaction force, the torque is zero: this is because the reaction force is applied exctly at the pivot point, so d = 0, and therefore the torque is zero.

Therefore:

- Weight: counterclockwise torque (I have assumed that the pencil is held at its right end)

- Reaction force: zero torque

8 0

3 years ago

Say you want to make a sling by swinging a mass M of 1.9 kg in a horizontal circle of radius 0.042 m, using a string of length 0

padilas [110]

Answer: T= 715 N

Explanation:

The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:

T = mv² / r

At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.

So the kinetic energy will be the following:

K = 1/2 m v² = 15. 0 J

Solving for v², and replacing in the expression for T:

T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N

3 0

3 years ago