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3 years ago

Volume can be measured in: A. cubic centimeters. B. centimeters. C. square centimeters.

Physics

2 answers:

Ad libitum [116K]3 years ago

8 0

Answer-

A. cubic centimeters.

Hope this helps!

podryga [215]3 years ago

6 0

Answer:

A. cubic centimeters.

Explanation:

Volume can be measured in:

A. cubic centimeters = volume

centimeter = length.

square centimeters = area

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A +35 µC point charge is placed 46 cm from an identical +35 µC charge. How much work would be required to move a +0.50 µC test c

Ira Lisetskai [31]

Answer:

512.5 mJ

Explanation:

Let the two identical charges be q = +35 µC and distance between them be r₁ = 46 cm. A charge q' = +0.50 µC located mid-point between them is at r₂ = 46 cm/2 = 23 cm = 0.23 m.

The electric potential at this point due to the two charges q is thus

V = kq/r₂ + kq/r₂

= 2kq/r₂

= 2 × 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C/0.23 m

= 630/0.23 × 10³ V

= 2739.13 × 10³ V

= 2.739 MV

When the charge q' is moved 12 cm closer to either of the two charges, its distance from each charge is now r₃ = r₂ + 12 cm = 23 cm + 12 = 35 cm = 0.35 m and r₄ = r₂ - 12 cm = 23 cm - 12 cm = 11 cm = 0.11 cm.

So, the new electric potential at this point is

V' = kq/r₃ + kq/r₄

= kq(1/r₃ + 1/r₄)

= 9 × 10⁹ Nm²/C² × 35 × 10⁻⁶ C(1/0.35 m + 1/0.11 m)

= 315 × 10³(2.857 + 9.091) V

= 315 × 10³ (11.948) V

= 3763.62 × 10³ V

= 3.764 MV

Now, the work done in moving the charge q' to the point 12 cm from either charge is

W = q'(V' - V)

= 0.5 × 10⁻⁶ C(3.764 MV - 2.739 MV)

= 0.5 × 10⁻⁶ C(1.025 × 10⁶) V

= 0.5125 J

= 512.5 mJ

8 0

3 years ago

A soap bubble, when illuminated with light of frequency 5.27 Hz × 1014 Hz, appears to be especially reflective. If it is surroun

denis23 [38]

Answer:

$1.07004\times 10^{-7}\ m$

Explanation:

$n_s$ = Refractive index of bubble = 1.33

f = Frequency of light = $5.27\times 10^{14}\ Hz$

c = Speed of light = $3\times 10^8\ m/s$

The wavelength of light is given by

$\lambda=\dfrac{2n_st}{m-\dfrac{1}{2}}$

Wavelength is also given by

$\lambda=\dfrac{c}{f}$

m = 1 for minimum thickness

$\dfrac{c}{f}=\dfrac{2n_st}{m-\dfrac{1}{2}}\\\Rightarrow t=\dfrac{m-\dfrac{1}{2}c}{2n_sf}\\\Rightarrow t=\dfrac{(1-\dfrac{1}{2})\times 3\times 10^8}{2\times 1.33\times 5.27\times 10^{14}}\\\Rightarrow t=1.07004\times 10^{-7}\ m$

The minimum thickness is $1.07004\times 10^{-7}\ m$

4 0

3 years ago

Now imagine a person dragging a 50 kg box along the ground with a rope, as

ANTONII [103]

Answer:

The coefficient of static friction between the box and floor is, μ = 0.061

Explanation:

Given data,

The mass of the box, m = 50 kg

The force exerted by the person, F = 50 N

The time period of motion, t = 10 s

The frictional force acting on the box, f = 30 N

The normal force on the box, η = mg

= 50 x 9.8

= 490 N

The coefficient of friction,

μ = f/ η

= 30 / 490

= 0.061

Hence, the coefficient of static friction between the box and floor is, μ = 0.061

7 0

4 years ago

Two immersion heaters, A and B, are both connected to a 120.0-V supply. Heater A can raise the temperature of 1.00 L of water fr

Inessa05 [86]

Answer:

Ratio of resistance of heater A to resistance of heater B is 5.80

Explanation:

Consider C be the specific heat of water, R₁ and R₂ be the resistance of heater A and heater B respectively.

Given:

Mass of water in heater A, m₁ = 1 L

Mass of water in heater B, m₂ = 5.80 L

Initial temperature, T₀ = 20 ⁰C

Final temperature, T₁ = 90 ⁰C

Time, t = 5 min

Amount of heat required to raise the temperature of water by heaters A and B are given by:

Q₁ = m₁C(T₁ - T₀) and

Q₂ = m₂C(T₁ - T₀)

Ratio of power used by both the heaters A and B is:

$\frac{P_{1} }{P_{2} } =\frac{Q_{1} }{t} \times\frac{t}{Q_{2} }$

Since, time t, temperature difference(T₁ - T₀) and specific heat C are same for both the heaters A and B. So, the above equation becomes:

$\frac{P_{1} }{P_{2} } =\frac{m_{1} }{m_{2} }$ ...(1)

The relation to determine electrical power for both heaters A and B are:

$P_{1}=\frac{V^{2} }{R_{1} }$ and

$P_{2}=\frac{V^{2} }{R_{2} }$

Here V is the voltage applied to both the heaters and is equal.

So, the ratio of electrical power of heaters is:

$\frac{P_{1} }{P_{2} } =\frac{R_{2} }{R_{1} }$ ....(2)

But according to the problem, the electrical power is converted into the thermal power. So,equation (1) and (2) are equal. Hence,

$\frac{m_{1} }{m_{2} } =\frac{R_{2} }{R_{1} }$

Substitute the suitable values in the above equation.

$\frac{1 }{5.80 } =\frac{R_{2} }{R_{1} }$

$\frac{R_{1} }{R_{2} }=5.80$

6 0

3 years ago

1. Find an example of electrical energy being converted to light energy.

Svetradugi [14.3K]

Solar energy is the example of electrical energy

6 0

4 years ago