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kipiarov [429]
2 years ago
6

A solid aluminum sphere of radius R has moment of inertia I about an axis through its center. What is the moment of inertia abou

t a central axis of a solid aluminum sphere of radius 2R
Physics
1 answer:
frutty [35]2 years ago
6 0

Answer:

I1 = 2/5 M1 R^2   for a sphere about its center

I2 = 2/5 M2 (2 R)^2 = 2/5 M2 R^ * 4 = 8/5 M2 R^2

Remember that M2 is greater than M1 by a factor 0f 2^3 = 8

Then I2 exceeds I1 by a factor of 32

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An electron is moving in the presence of an electric field of 400 N/C. What force does the electron experience?
yanalaym [24]

Answer:

Explanation:

Given

Electric Field strength E=400\ N/C

charge of electron q=1.6\times 10^{-19}\ C

Force on the charge particle is given by

F=qE

F=1.6\times 10^{-19}\times 400

F=640\times 10^{-19}\ N

but this force will be acting in the direction opposite to the direction of Electric field because electron is negatively charged                                            

8 0
3 years ago
Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res
damaskus [11]

Answer:

Same, the electric fields point in opposite directions and therefore cancel at some midpoint.

Explanation:

The Electric field net at some point between them is zero, only if they point in opposite direction (they cancel to the each other). In order the electric fields  have opposite direction, at some point between the bees , the bees must have the same sign of electric charge

3 0
3 years ago
Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium
ss7ja [257]

Answer:

The number of atoms are 1.86\times10^{8}.

Explanation:

Given that,

Diameter D = 1.40\times10^{2}\ pm

D=1.40\times10^{2}\times10^{-12}\ m

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}

Number\of\atoms =1.86\times10^{8}

Hence, The number of atoms are 1.86\times10^{8}.

8 0
2 years ago
Batman (95kg) is standing on top of a 50m high building looking out over the city of Gotham. Given that he uses the potential en
Oksanka [162]

Answer:

47 kJoules (kJ)

Explanation:

Potential enegy on Earth is given by the relationship:

P.E. = mgh, where m is mass, g is the acceleration due to Earth's gravity, and h is height. Since we are given metric values, we will look for an answer that is consistent with Joules, the metric measure of energy. 1 Joule is defined as 1 kg*m^2/s^2, so we wnat units of kg, m, and sec.

We are given:

m = 95kg

h = 50 meters

Earth's gravity, g is 9.8 m/s^2

Enter the data:

P.E. = mgh

P.E. = (95kg)(9.8m/s^2)(50m)

P.E. = 46550 kg*m^2/s^2 or 46550 Joules(J)

Since we only have 2 sig figs, and since 1kJ =- 1000J

We can state the potential energy is 47kJ.

Spiderman has 47kJ of potential energy for the start of any dive back to Earth. [He needed that same amount of energy to reach that height, but we don't know from where it came. A jump, helicopter, beamed up by Scotty, or tossed up by Doctor Octopus.]

3 0
1 year ago
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