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Mashutka [201]
3 years ago
10

As you (increase, decrease) in altitude, air pressure decreases.

Physics
1 answer:
Bingel [31]3 years ago
5 0

Answer:

as you increase in altitude, air pressure decreases

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Part III: If a mouse and an elephant both run with the same kinetic energy, can you say which is running faster? Use the space b
Verdich [7]

Answer:

The mouse runs faster to have the same kinetic energy as the elephant.

Explanation:

Note from the equation given, mass (m) is directly proportional to KE. This means an elephant with more mass will have more KE, therefore, for the mouse to compensate, it has to run faster because its KE is smaller because of its small mass. If both run at the same speed, the elephant would have thousands of times more kinetic energy than the mouse. So the mouse has to run faster so that its speed compansates for its smaller weight.

3 0
3 years ago
A person drives a car around a circular road with a constant speed of 20 m/s. The
ale4655 [162]

Answer:

16 m/s^2

Explanation:

acceleration tangential = (v^2)/r

a=400/25

a=16 m/s^2

Side note: next time, be more specific when asking about acceleration in circular motion. There's more than one type! Example:

angular acceleration=acceleration tangential/r

angular acc.=16/25

angular acc.=0.64 rad/s^2

5 0
3 years ago
Read 2 more answers
Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo
djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for  the force of static friction:

f_s = \mu_s N --- (1)

where;

f_s = static friction force

\mu_s = coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

N = ma_{min}

From equation (1)

f_s = \mu_s ma_{min}

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

F = f_s

mg = \mu_s ma_{min}

a_{min} = \dfrac{mg }{ \mu_s m}

a_{min} = \dfrac{g }{ \mu_s }

where;

\mu_s = 0.383 and g = 9.8 m/s²

a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }

\mathbf{a_{min}= 25.59 \ m/s^2}

3 0
3 years ago
A small ball of mass 2.00 kilograms is moving at a velocity 1.50 meters/second. It hits a larger, stationary ball of mass 5.00 k
rewona [7]

The kinetic energy of the small ball before the collision is

                             KE  =  (1/2) (mass) (speed)²

                                     = (1/2) (2 kg) (1.5 m/s)

                                     =    (1 kg)  (2.25 m²/s²)

                                     =        2.25 joules.

Now is a good time to review the Law of Conservation of Energy:

                     Energy is never created or destroyed. 
                     If it seems that some energy disappeared,
                     it actually had to go somewhere.
                     And if it seems like some energy magically appeared,
                     it actually had to come from somewhere.

The small ball has 2.25 joules of kinetic energy before the collision.
If the small ball doesn't have a jet engine on it or a hamster inside,
and does not stop briefly to eat spinach, then there won't be any
more kinetic energy than that after the collision.  The large ball
and the small ball will just have to share the same 2.25 joules.

3 0
3 years ago
Incident rays parallel to the principle axis of a concave mirror will reflect _____. parallel to the principle axis through the
Svetradugi [14.3K]
I’m not sure if its correct but I think it’s focal Ray point


For concave mirrors, some generalizations can be made to simplify ray construction. They are: An incident ray traveling parallel to the principal axis will reflect and pass through the focal point. An incident ray traveling through the focal point will reflect and travel parallel to the principal axis.
8 0
4 years ago
Read 2 more answers
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