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Oduvanchick [21]
3 years ago
5

Calculate the force experienced by a positive test charge of 8.1 × 10-9 coulombs if the electric field strength is 9.4 × 107 new

tons/coulomb.
Physics
1 answer:
luda_lava [24]3 years ago
4 0
We can answer this problem by dimensional analysis. As we can see, the units of the electric field strength is newtons per coulomb. If we multiply a value with units of coulomb we will obtain newtons which is a units of force. 

Force = 9.4x10^7 N / C (8.1 x10^-9 C)
Force = 0.7614 N
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Two narrow, parallel slits separated by 0.85 mm are illuminated by 600 nm light, and the viewing screen is 2.8 m away from the s
AURORKA [14]

Answer:

Phase difference = pi/4 radians

Explanation:

Given:

- The wavelength of incident light λ = 600 nm

- The split separation d = 0.85 mm

- Distance of screen from split plane L = 2.8 m

Find:

What is the phase difference between the two interfering waves on a screen, at a point 2.5 mm from the central bright fringe?

Solution:

- The phase difference can be evaluated by determining the type of interference that occurs at point y = 2.5 mm above central order. We will use the derived results from Young's double slit experiment.

                                  sin ( Q ) = m*λ /d  

                                  m = d*sin(Q) / λ

- Where, m is the order number and angle Q is the angle for mth order of fringe from central bright fringe.

                                  r = sqrt ( L^2 + 0.0025^ )

Where, r is the distance from split to the interference bright fringe.

                                  r = sqrt(2.8^ + 0.0025^) = 2.8

                                  sin(Q) = 0.0025 / 2.8

Hence.                        m = 0.00085*0.0025 / 2.8*(600*10^-9)

                                   m = 1.26

- We know that constructive interference would occurred at m = 1 and destructive interference @ m = 1.5. They have a phase difference of pi/2 radians.

- The order number lies in between constructive and destructive interference i.e m ≈ 1.25 then the corresponding phase difference = 0.5*(pi/2).

Answer:                  Phase difference = pi/4 radians

6 0
3 years ago
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

5 0
2 years ago
Read 2 more answers
What is the distance traversed by the particle between 0 seconds and 6 seconds?
ad-work [718]

Answer:

I dont know to be honest

Explanation:

i dont know to be honest

5 0
3 years ago
At a given time of​ day, the ratio of the height of an object to the length of its shadow is the same for all objects. If a 4​-f
katrin2010 [14]

Answer:35.2 ft

Explanation:

Given

height of stick =4 ft

shadow length =2.8 ft

Angle of elevation of sun is

tan\theta =\frac{4}{2.8}

let the height of tree be h

as \thetawill remain same thus

tan\theta =\frac{h}{24.64}

\frac{4}{2.8}=\frac{h}{24.64}

h=35.2 ft

8 0
3 years ago
Which characteristic do all minerals have in common?
Korolek [52]

Answer:

Five Characteristics of a Mineral

Minerals Are Natural. You must find minerals in nature; substances concocted in laboratories don't qualify. ...

Minerals Are Inorganic. ...

Minerals Are Solids. ...

Definite Chemical Composition. ...

Crystalline Structure.

These are just some examples.

Explanation:

7 0
2 years ago
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