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AleksAgata [21]
3 years ago
7

Read the scenario. A car starts 10 m north of a reference point. It moves at a constant velocity over the next 5 s, reaching a p

osition of 10 m south of the reference point. What is the car’s average velocity? 2 m/s south 4 m/s north 0 m/s 4 m/s south
Physics
1 answer:
Gennadij [26K]3 years ago
4 0

Answer:

4 m/s south

Explanation:

This is a pretty easy one.

Assume that the car has a reference point ragged x

We also know that the car starts moving at a point 10 m North of X. It moves steadily at a uniform velocity and ended at a position 10 m south of X.

If x is the reference point, we can infer that the car moves a total distance of, 10 m + 10 m.

Thus, the total distance moved by the car is 20 m.

It is also stated that it achieves that distance in 5 seconds.

Velocity is defined as the ratio of distance with respect to the tome taken, i.e V = d/t

Then, the velocity is

V = 20 / 5 = 4 m/s

Considering the fact that the car left and moved towards the south. We can say that it moved 4 m/s South.

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heterogeneous and homogeneous

Explanation:

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An upward force act on a proton as it moves with a speed of 2.0 x 10^5 meters/seconds through a magnetic field of 8.5 x 10^2
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Force on a moving charge is given by formula

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here we know that this force will be maximum when velocity is perpendicular to magnetic field

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here we know that

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q = 1.6 \times 10^{-19} C

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now we have

F = (1.6 \times 10^{-19})(2 \times 10^5)(8.5 \times 10^2)

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3 years ago
Which sound waves in the electromagnetic spectrum are considered low energy waves
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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.

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Answer:

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Explanation:

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