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ruslelena [56]
1 year ago
9

A particular planet has a moment of inertia of 9.74 × 1037 kg ⋅ m2 and a mass of 5.98 × 1024 kg. Based on these values, what is

the planet's radius? Hint: Because planets are the shape of a sphere, the moment of inertia is I = (2/5)mr2.
A) 6.38 x 106 m
B) 2.55 × 106 m
C) 6.52 × 1012 m
D) 4.07 × 1013m
Physics
1 answer:
malfutka [58]1 year ago
4 0

Answer:  A) 6.38(10)^{6} m

Explanation:

The equation for the moment of inertia I of a sphere is:

I=\frac{2}{5}mr^{2} (1)

Where:

I=9.74(10)^{37}kg m^{2} is the moment of inertia of the planet (assumed with the shape of a sphere)

m=5.98(10)^{24}kg is the mass of the planet

r is the radius of the planet

Isolating r from (1):

r=\sqrt{\frac{5I}{2m}} (2)

Solving:

r=\sqrt{\frac{5(9.74(10)^{37}kg m^{2})}{2(5.98(10)^{24}kg)}} (3)

Finally:

r=6381149.077m \approx 6.38(10)^{6} m

Therefore, the correct option is A.

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<em>ANSWER: D</em>
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2 years ago
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Suppose an electrical wire is replaced with one having every linear dimension doubled (i.e., the length and radius have twice th
Flauer [41]

Answer:

The new resistance becomes half of the initial resistance.

Explanation:

The resistance of a wire is given by :

R=\dfrac{\rho L}{A}

\rho = resistivity of material

L and A are linear dimension

If the electrical wire is replaced with one having every linear dimension doubled i.e. l' = 2l and r' = 2r

New resistance of wire is given by :

R'=\dfrac{\rho L'}{A'}

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R'=\dfrac{1}{2}\dfrac{\rho L}{A}

R'=\dfrac{1}{2}R

The new resistance becomes half of the initial resistance. Hence, this is the required solution.

4 0
2 years ago
One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its hi
Aneli [31]

Answer:

Explanation:

Given

initial speed(u)=5 m/s

Final speed(v)=4 m/s

Distance traveled=3 m

using equation of motion

v^2-u^2=2as

4^2-5^2=2(a)(3)

a=\frac{-3}{2}=-1.5 m/s^2

after this its final velocity will be zero

v^2-u^2=2as

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s=5.33 m

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4 0
1 year ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

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2 years ago
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Answer:

6.67ft/s^2

Explanation:

We are given that

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Final velocity,v=38ft/s

Time=t=3 s

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We know that

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Average acceleration,a=\frac{20}{3}ft/s^2

Average acceleration,a=6.67ft/s^2

Hence, the average acceleration=6.67ft/s^2

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