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monitta
3 years ago
8

What the answer to the three of them ill mark brainlest

Physics
1 answer:
natali 33 [55]3 years ago
6 0

lhormmia  masng ies igentetel

Explanation:

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If your car tachometer says your engine is moving at 1200 RPM, then what is it's angular velocity in Rad/sec?
Ostrovityanka [42]

Answer:

125.66 R/s

Explanation:

First    1200  r / min = 20 r/sec

20 r/s  *  2pi Radians / r = 40 pi Radians / sec = 125.66 R/s

3 0
2 years ago
he acceleration due to gravity on the surface of Mars is about one third the acceleration due to gravity on Earth’s surface. The
Ahat [919]

1/3 the weight than it is on earth

5 0
4 years ago
Please help me! this is timed!
Arada [10]

Answer:

Radiation

Explanation:

Radiation refers to the emission of energy in rays or waves

8 0
3 years ago
A baseball pitcher loosens up his pitching arm. He tosses a 0.20-kg ball using only the rotation of his forearm, 0.28 m in lengt
gayaneshka [121]

Answer:

Moment of Inertia, I = 0.016 kgm²

Explanation:

Mass of the ball, m = 0.20 kg

Length of the pitcher's arm, l = 0.28

Radius of the circular arc, r = 0.28 m

Moment of Inertia is given by the formula:

I = mr²

I = 0.20 * 0.28²

I = 0.20 * 0.0784

I = 0.01568

I = 0.016 kgm²

5 0
4 years ago
A solid conducting sphere with radius RR that carries positive charge QQ is concentric with a very thin insulating shell of radi
svetlana [45]

Answer:

E=0 at r < R;

E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}} at 2R > r > R;

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}} at r >= 2R

Explanation:

Since we have a spherically symmetric system of charged bodies, the best approach is to use Guass' Theorem which is given by,

\int {E} \, dA=\frac{Q_{enclosed}}{\epsilon} (integral over a closed surface)

where,

E = Electric field

Q_{enclosed} = charged enclosed within the closed surface

\epsilon = permittivity of free space

Now, looking at the system we can say that a sphere(concentric with the conducting and non-conducting spheres) would be the best choice of a Gaussian surface. Let the radius of the sphere be r .

at r < R,

Q_{enclosed} = 0 and hence E = 0 (since the sphere is conducting, all the charges get repelled towards the surface)

at 2R > r > R,

Q_{enclosed} = Q,

therefore,

E\times4\pi r^{2}=\frac{Q_{enclosed}}{\epsilon}      

(Since the system is spherically symmetric, E is constant at any given r and so we have taken it out of the integral. Also, the surface integral of a sphere gives us the area of a sphere which is equal to 4\pi r^{2})

or, E=\frac{1}{4\pi\epsilon}\frac{Q}{r^{2}}

at r >= 2R

Q_{enclosed} = 2Q

Hence, by similar calculations, we get,

E=\frac{1}{4\pi\epsilon} \frac{2Q}{r^{2}}

4 0
4 years ago
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