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3 years ago

What are the examples of transparent medium?

Physics

1 answer:

Dennis_Churaev [7]3 years ago

3 0

Answer:

fog and mist are the examples

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Giving me the points are enough <br>

N76 [4]

Answer:

the product of mass and velocity

....in my syllabus

7 0

3 years ago

A point charge q = +4.50 nC moves through a potential difference ΔV = Vf − Vi = +27.0 V. What is the change in the electric pote

olganol [36]

Change in electric potential energy: 121.5 nJ

Explanation:

For a charged particle moving in an electric field, the change in electric potential energy of the particle is given by

$\Delta U = q \Delta V$

where:

q is the charge of the particle

$\Delta V$ is the potential difference between the initial and final position of the particle

For the point charge in this problem, we have:

$q=+4.50 nC$ is the charge

$\Delta V=+27.0 V$ is the potential difference

Therefore, the change in electric potential energy is

$\Delta U=(+4.50)(+27.0)=121.5 nJ$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0

3 years ago

What does it mean for their to be a net force on an object vs no net force? Which one is described as balanced forces and which

STatiana [176]

It’s like the force put against something or someone✨

3 0

2 years ago

Lake Baikal in Siberia has a maximum depth of 1642 m. What is the water

Veseljchak [2.6K]

Explanation:

Fluid gauge pressure is:

P = ρgh

where ρ is the fluid density and h is the depth of the fluid.

P = (1000 kg/m³) (9.8 m/s²) (1642 m)

P = 16,091,600 Pa

Rounded to four significant figures, the gauge pressure is 16.09 MPa.

3 0

3 years ago

A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0

3 years ago