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kap26 [50]
3 years ago
9

What has a larger capacitance, an aluminum sphere with a 10 cm diameter or one with a 100 cm diameter? Question 16 options: 10 c

m 100 cm They are both the same. cannot be determined
Physics
1 answer:
creativ13 [48]3 years ago
3 0

\boxed{\sf C=\dfrac{Q}{V}}

But

\boxed{\sf \Delta V_{R_2\to R_1}={\displaystyle{\int}^{R_1}_{R_2}}dV=-{\displaystyle{\int}^{R_1}_{R_2}}\dfrac{kQ}{r^2}dR}

  • Hence higher the radius lower the voltage
  • Lower the voltage higher the capacitance .

<h3>100cm diameter having aluminium sphere has a larger capacitance</h3>
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Bond [772]
A = .3*g = 2.94 m/s² 

<span>t = v/a = 9/2.94 = 3.061 sec </span>

<span>W = E/t = ½mv²/t = ½*40*9²/3.061 = 529.2 watts</span>
4 0
3 years ago
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cable
kompoz [17]

Answer:

4.44s

Explanation:

A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables

since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation

T=2π√L/g

g=acceleration due to gravity which is 9.81m/s2

the length of the supporting cable is 4.9m

T the period

period is the time required to make a complete oscillation

T=2*π√4.9/9.81

T=2*π*0.706

T=4.44s

4.44s

5 0
3 years ago
Determine whether the variable is qualitative or quantitative. Favorite sport. Is the Variable Qualitative or Quantitative?
Over [174]
The variable is qualitative, 
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4 0
3 years ago
4. A 1200 kg car traveling North at 20.0 m/s collides with a 1400 kg car traveling South at 22.0 m/s. The two
Dvinal [7]

Answer:-2.61 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum p_{o} must be equal to the final momentum p_{f}:

p_{o}=p_{f} (1)

Where:

p_{o}=mV_{o}+MU_{o} (2)

p_{f}=(m+M)V_{f} (3)

m=1200 kg is the mass of the first car

V_{o}=20 m/s is the velocity of the first car, to the North

M=1400 kg is the mass of the second car

U_{o}=-22 m/s is the mass of the second car, to the South

V_{f} is the final velocity of both cars after the collision

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1200 kg)(20 m/s)+(1400 kg)(-22 m/s)}{1200 kg+1400 kg} (6)

Finally:

V_{f}=-2.61 m/s (7) This is the resulting velocity of the wreckage, to the south

7 0
4 years ago
If the beam carries 1015 electrons per second and is accelerated by a 350 kV source, find the current and power in the beam.
Y_Kistochka [10]

To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,

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Here q is Flow of electric charge in one second and N the number of electron flow per second.

A the same time the power is described as the applied voltage for the current.

P = VI

We know the charge of electron, q = 1.602 * 10^{-19} Coulombs, then the current is

I = (1.602*10^{-19})(10^{15})

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And the power in the Beam is

P = VI

P = (350*10^3)(0.1602)

P = 0.05607 Watts

3 0
3 years ago
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