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VikaD [51]
3 years ago
6

CAN AIR MAKE SHADOWS?

Physics
2 answers:
Tems11 [23]3 years ago
3 0

Answer:

No because you cannot see air so therefore it cannot make shadows

Explanation:

anyanavicka [17]3 years ago
3 0

Answer:

NO, but it can move things that have shadows which makes the shadow move.

Like tree branches for example when the wind blows the branches move which makes their shadow move also.

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An unbalanced force gives a 2.00 kg mass an acceleration of 5.00 m/s? What is the force applied to the object?​
valentinak56 [21]

Answer:

10N

Explanation:

Equation: ΣF = ma

Fapp = ma

Fapp = (2kg)(5m/s^2)    (im guessing you mean 5.00 m/s^2 not m/s)

Fapp = 10*kg*m/s^2

Fapp = 10N

5 0
2 years ago
What does an electric circuit mean?​
nikitadnepr [17]

Answer: An electric circuit includes a device that gives energy to the charged particles constituting the current, such as a battery or a generator; devices that use current, such as lamps, electric motors, or computers; and the connecting wires or transmission lines.

Explanation: Hope this helped! <3

6 0
3 years ago
Which statement best explains why an object
slavikrds [6]
D, Mercury as a weaker gravitational pull! Due to mercury being farther from the sun and it being a smaller planet it has a weaker pull
4 0
3 years ago
Read 2 more answers
6^5/2^3= what is the answer please​
kogti [31]

Answer:

972

Explanation:

6^5/2^3

=7776/8

=972

6 0
3 years ago
Read 2 more answers
A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box
forsale [732]
We need to see what forces act on the box:

In the x direction:

Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

In the y direction:

N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force. 

From N-Gcosα=0 we get: 

N=Gcosα, we will need this for the force of friction.

Now to solve for Fh:

Fh=ma + Ff + Gsinα,

Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.
G=mg, where m is the mass of the box and g=9.81 m/s²

Fh=ma + μmgcosα+mgsinα

Now we plug in the numbers and get:

Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
8 0
3 years ago
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