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3 years ago

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Two spheres of the same size and mass roll down an incline. Sphere A is hollow and Sphere B is solid with uniform density. Which

reaches the bottom of the incline first?
Make sure to EXPLAIN your answer!

(a) Sphere A
(b) Sphere B
(c) At the same time.

1 answer:

sweet [91]3 years ago

8 0

If the spheres are have to have the same and the same size, it is expected that they will have the same surface area. The only factor that affects the descent of the object through gravity on Earth is the surface area that is exposed to the resistance of air. For this case, it is expected that they will reach the bottom of the inclined plane together. The answer is C.

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Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

3 years ago

Using an unmanned rocket to visit the space station requires 85.2 trillion BTU of energy. The best fuel for the mission will hav

Illusion [34]

We shall convert all of the densities to lbs/gal, so the product of
BTU/lbs and lbs/gal gives us the basis of comparison, which was "ratio of energy to volume".
grams / ml x 1 lbs/454 grams → 1 lbs/ 454 ml
1 lbs/454 ml x 3785.41 ml/gal → 3785.41 lbs/454gal
Conversion of g/ml = 8.34 lbs/gal
Looking at each fuel:

Kerosene:
18,500 x (8.34 x 0.82) = 126,517 BTU/gal

Gasoline:
20,900 x (8.34 x 0.737) = 128,463 BTU/gal

Ethanol:
11,500 x (8.34 x 0.789) = 75,673 BTU/gal

Hydrogen:
61,000 x (8.34 x 0.071) = 36,120 BTU/gal

The best fuel in terms of energy to volume ratio is Gasoline.
Gallons required:
BTU needed / BTU per gallon
= 85.2 x 10⁹ / 128,463
= 6.6 x 10⁵ gallons

5 0

3 years ago

Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur

devlian [24]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

2 y = 0.025 m

y = 0.0125 m

For circular aperture

$sin \theta = 1.22\dfrac{\lambda}{d}$

using small angle approximation

$\theta = \dfrac{y}{D}$

now,

$\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}$

$y = 1.22\dfrac{\lambda\ D}{d}$

$d = 1.22\dfrac{\lambda\ D}{y}$

$d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}$

d =0.247 x 10⁻³ m

d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

5 0

3 years ago

A doppler effect occurs when a source of sound moves. True or False

Shtirlitz [24]

<h2>Answer: True </h2>

The <u>Doppler effect</u> refers to the change in a wave perceived frequency when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other.

In other words, it is the variation of the frequency of a wave due to the relative movement of the source of the wave with respect to its receiver.

It should be noted that this effect bears its name in honor of the Austrian physicist <u>Christian Andreas Doppler</u>, who in 1842 proposed the existence of this effect for the case of light in the stars. Another important aspect is that the effect occurs in all waves (including light and sound). However, it is more noticeable to humans with sound waves.

4 0

3 years ago

Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren

nikklg [1K]

Answer:

$4.77\ \text{A}$

Explanation:

F = Magnetic force = 4.11 N

$I_n$ = Net current

$I_2$ = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

$\theta$ = Angle between current and magnetic field = $65^{\circ}$

$l$ = Length of wires = 2.64 m

$I$ = Current in the other wire

Magnetic force is given by

$F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}$

Net current is given by

$I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}$

The current I is $4.77\ \text{A}$ .

8 0

3 years ago