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Andrei [34K]
3 years ago
15

"two planets have the same mass, but planet a has 3 times the radius of planet

Physics
1 answer:
lyudmila [28]3 years ago
4 0
The gravitational acceleration on the surface of planet A is:
g_A =  \frac{GM_A}{r_A^2}
where G is the gravitational constant, M_A is the mass of planet A and r_A its radius.

Similarly, the gravitational acceleration on the surface of planet B is:
g_B = \frac{GM_B}{r_B^2}

The ratio between the gravitational acceleration on planet A and B becomes:
\frac{g_A}{g_B}= \frac{GM_A / r_A^2}{GM_B/r_B^2}  = \frac{M_A r_B^2}{M_B r_A^2}

The problem says that the two masses are equal: M_A = M_B while planet A has 3 times the radius of planet B: r_A = 3 r_B. Substituting into the ratio, we get:
\frac{g_A}{g_B} =  \frac{M_B r_B^2}{M_B (3 r_B)^2} =  \frac{1}{9}

so, gravity on planet B is 9 times stronger than planet A.
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A thin walled spherical shell is rolling on a surface. What
ExtremeBDS [4]

Answer:

=\frac{1/3}{5/6} = 0.4

Explanation:

Moment of inertia of given shell= \frac{2}{3} MR^2

where

M represent sphere mass

R -sphere radius

we know linear speed is given as v = r\omega

translational K.E = \frac{1}{2} mv^2 = \frac{1}{2} m(r\omega)^2

rotational K.E = \frac{1}{2} I \omega^2 = \frac{1}{2} \frac{2}{3} MR^2 \omega^2

total kinetic energy will be

K.E = \frac{1}{2} m(r\omega)^2 + \frac{1}{2} \frac{2}{3} MR^2 \omega^2

K.E =\frac{5}{6} MR^2 \omega^2

fraction of rotaional to total K.E

=\frac{1/3}{5/6} = 0.4

8 0
3 years ago
A long wire carrying a 5.0 A current perpendicular to the xy-plane intersects the x-axis at x= - 2.0 cm . A second, parallel wir
mario62 [17]

Answer:

a . 0.35cm

b.  11.33cm

Explanation:

a. Given both currents are in the same direction, the null point lies in between them. Let x be distance of N from first wire, then distance from 2nd wire is 4-x

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in between the wires:

\frac{\mu_oi_1}{2\pi x}=\frac{\mu_oi_2}{2\pi(4-x)}\\\\5/x=\frac{3.5}{4-x}\\\\x=2.35cm\\\\N=2.35-2=0.35cm

Hence, for currents in same direction, the point is 0.35cm

b. Given both currents flow in opposite directions, the null point lies on the other side.

#For the magnetic fields to be zero,the fields of both wires should be equal and opposite.They are only opposite in outside the wires:

Let x be distance of N from first wire, then distance from 2nd wire is 4+x:

\frac{\mu_oi_1}{2\pi(4+ x)}=\frac{\mu_oi_2}{2\pi x}\\\\5/(4+x)=\frac{3.5}{x}\\\\x=9.33cm\\\\N=9.33+2=11.33cm

Hence, if currents are in opposite directions the point on x-axis is 11.33cm

8 0
3 years ago
Current that moves in one direction from negative to positive. May be created by a battery. Is generally NOT found in U.S. elect
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That's "<em>DC</em>" . . . Direct Current .

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3 years ago
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According to Newton's Law of Cooling, if a body with temperature T 1 is placed in surroundings with temperature T 0, different f
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T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F

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An electric grinder uses a grinding wheel
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(1500 rev/min)(min / 60 s) / (3.0 s) = 8.33 rev/s² 

<span>(B) </span>
<span>(1/2)(8.33 rev/s²)(3.0 s)² = 37.5 rev </span>

<span>(C) </span>
<span>(1500 rev/min)(min / 60 s)[2π(0.12 m) / rev] = 18.8 m/s</span>
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2 years ago
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