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marissa [1.9K]
3 years ago
11

How many electrons have been removed from a positively charged electroscope if it has a net charge of 6x10-11?

Physics
1 answer:
oksano4ka [1.4K]3 years ago
5 0

Answer:

is it the ans you are looking for

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A sample of an ideal gas is in a tank of constant volume. The sample absorbs heat energy so that its temperature changes from 33
Anuta_ua [19.1K]

Answer:

\frac{v_{2}}{v_{1}}=2.

Explanation:

The average kinetic energy per molecule of a ideal gas is given by:

\bar{K}=\frac{3k_{B}T}{2}

Now, we know that \bar{K} = (1/2)m\bar{v}^{2}

Before the absorption we have:

(1/2)m\bar{v_{1}}^{2}=\frac{3k_{B}T_{1}}{2} (1)

After the absorption,

(1/2)m\bar{v_{2}}^{2}=\frac{3k_{B}T_{2}}{2} (2)

If we want the ratio of v2/v1, let's divide the equation (2) by the equation (1)

\frac{v_{2}^{2}}{v_{1}^{2}}=\frac{T_{2}}{T_{1}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{T_{2}}{T_{1}}}

\frac{v_{2}}{v_{1}}=\sqrt{\frac{1340}{335}}

\frac{v_{2}}{v_{1}}=\sqrt{4}

Therefore the ratio will be \frac{v_{2}}{v_{1}}=2

I hope it helps you!

4 0
4 years ago
Read 2 more answers
1. A car travels a distance of 200 m in 4 seconds. What is the velocity of the car?
dybincka [34]

Answer:

velocity = distance / time taken

= 200/4

= 50 m/s

is the correct answer

3 0
3 years ago
A stretched string has a mass per unit length of 4.87 g/cm and a tension of 16.7 N. A sinusoidal wave on this string has an ampl
Burka [1]

Answer:

Explanation:

Given that,

Mass per unit length is

μ = 4.87g/cm

μ=4.87g/cm × 1kg/1000g × 100cm/m

μ = 0.487kg/m

Tension

τ = 16.7N

Amplitude

A = 0.101mm

Frequency

f = 71 Hz

The wave is traveling in the negative direction

Given the wave form

y(x,t) = ym• Sin(kx + ωt)

A. Find ym?

ym is the amplitude of the waveform and it is given as

ym = A = 0.101mm

ym = 0.101mm

B. Find k?

k is the wavenumber and it can be determined using

k = 2π / λ

Then, we need to calculate the wavelength λ using

V = fλ

Then, λ = V/f

We have the frequency but we don't have the velocity, then we need to calculate the velocity using

v = √(τ/μ)

v = √(16.7/0.487)

v = 34.29

v = 5.86 m/s

Then, we can know the wavelength

λ = V/f = 5.86 / 71

λ = 0.0825 m

So, we can know the wavenumber

k = 2π/λ

k = 2π / 0.0825

k = 76.18 rad/m

C. Find ω?

This is the angular frequency and it can be determined using

ω = 2πf

ω = 2π × 71

ω = +446.11 rad/s

D. The angular frequency is positive (+) because the direction of propagation of wave is in the negative direction of x

5 0
3 years ago
4.
galina1969 [7]

Answer:

72%

Calculating Efficiency:

The amount of work obtained from a machine is always less than the amount of work put into it. Thisis because some of the work is lost due to friction. The efficiency of a machine can be calculated usingthe following formula:Percent Efficiency = (work output-Machine / work input-You) x 100%

8 0
2 years ago
Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
4 years ago
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