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Svet_ta [14]
3 years ago
11

What do we call air in the motion​

Physics
2 answers:
Vaselesa [24]3 years ago
5 0

Answer:

it is wind in geography you will see it air in motion is wind

r-ruslan [8.4K]3 years ago
3 0
Answer: i think its wind
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A uniform rod of mass M and length L can pivot freely at one end. Initially, the rod is oriented vertically above the pivot, in
Leya [2.2K]

Answer:

The speed of its center of mass =\sqrt{\frac{3}{2}gL}

Explanation:

Consider the potential energy at the level of center of mass of rod below the pivot=0

Mass of uniform rod=M

Length of rod=L

The rotational inertia about the end of a uniform rod=\frac{1}{3}ML^2

Kinetic energy at the level of center of mass of rod below the pivot=\frac{1}{2}I\omega^2

Kinetic energy at the level of center of mass of rod above the pivot=0

Potential energy at the level of center of mass of rod above the pivot=mgh

We have to find the center of mass ( in terms of g and L).

According to conservation of law of energy

Initial P.E+Initial K.E=Final P.E+Final K.E

mgh+0=0+\frac{1}{2} I\omega^2

Where K.E=\frac{1}{2} I\omega^2

I=Moment of inertia

\omega=Angular velocity

Substitute the values then we get

MgL=\frac{1}{2}\times \frac{1}{3}ML^2\omega^2

\omega^2=\frac{6g}{L}

Now, we know that \omega=\frac{v}{r}, r=\frac{L}{2}

Substitute the values then we get

\frac{v^2}{(\frac{L}{2})^2}=\frac{6g}{L}

\frac{v^24}{L^2}=\frac{6g}{L}

v^2=\frac{6g\times L^2}{4L}

v^2=\frac{3gL}{2}

v=\sqrt{\frac{3}{2}gL}

Hence, the speed of its center of mass =\sqrt{\frac{3}{2}gL}

4 0
4 years ago
A 7.4 A current is set up in a circuit for 7.8 min by a rechargeable battery with a 3.0 V emf. By how much is the chemical energ
bearhunter [10]

Answer:

10389.6 J

Explanation:

Power is the rate of doing work with respect to time, its S.I unit is in watts but it can also be expressed in J/s. Power is calculated using the formula:

Power=\frac{energy}{time}

Power is also the rate at which energy is used per second.

Energy is the capacity to do work and it is measured in joules (J).

Power = current × voltage = 7.4 A × 3 V = 22.2 W

Energy=power*time\\\\time=7.8\ min=7.8*60\ s=468\ s\\\\Energy=22.2*468\\\\Energy=10389.6\ J

8 0
3 years ago
Problem 5 You are playing a game where you drop a coin into a water tank and try to land it on a target. You often find this gam
Dvinal [7]

Answer:

Option D: 1.5in in front of the target

Explanation:

The object distance is y= 6in.

Because the surface is flat, the radius of curvature is infinity .

The incident index is n_i=\frac{4}{3} and the transmitted index is n_t= 1.

The single interface equation is \frac{n_i}{y}+\frac{n_t}{y^i}=\frac{n_t-n_i}{r}

Substituting the quantities given in the problem,

\frac{\frac{4}{3}}{6in}+\frac{1}{y^i}= 0

The image distance is then y^i=-\frac{18}{4}in =-4.5in

Therefore, the coin falls 1.5in in front of the target

6 0
3 years ago
Physics Question - Matching
KiRa [710]
1) G

2) E

3) D

4) I

5) J

6) C

7) H

8) F

9) B

10) A

I think... i am not 100% sure....
7 0
4 years ago
Read 2 more answers
A forward honzontal force of 50 N is applied to a crate. A second horizontal force o
True [87]

Answer:

Explanation:

magnitude: 180-50=130N

Direction: in the direction same as the second horizontal force

4 0
3 years ago
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