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xenn [34]
4 years ago
14

A 10 cm long tensile specimen with a yield stress of 350 MPa is pull to a stress of 300 MPa and then the load is released. What

is the length of the specimen after the load is released?
Physics
1 answer:
Advocard [28]4 years ago
3 0

Answer:

The length of the specimen after the load is released is 11.67 cm

Explanation:

Given;

yield stress, Y = 350 MPa

ultimate tensile stress, T = 300 MPa

Elongation factor, e = yield stress, Y / ultimate tensile stress, T

Elongation factor, e = 350 Mpa / 300 Mpa

Elongation factor, e = 1.1667

New length of the specimen = 1.1667 x 10 cm = 11.67 cm

Therefore, when the load is released from 10 cm long tensile specimen, the length of the specimen becomes 11.67 cm

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9. According to statistical theory, 68% of your measurements of time should fall within the range of and of therefore about 3 of
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Yes, this is a random error generating because of statistical constraint. We only have finite number of data points. As per this, if we plot our observation we will get a gaussian (inverse bell ) shaped curve with mean equal to central value.

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3 years ago
1.A Radio station broadcasts modern song on medium wave 350 Hz every day at ten o’clock in the morning. The velocity of radio wa
love history [14]

Answer:

ans \:  = \boxed{{4.8 \times 10}^{ - 4}  Hz}

Explanation:

given \to \\  f_{r} = 350 \:  \\ v_{r} =  {3 \times 10}^{8}  \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{r} =  \frac{v_{r}}{f_{r}}   =  \frac{3 \times 10^{8} }{350}   =  \boxed{857,142.85714 \: m}\\ therefore \to \\ given \to \\  f_{w} = water \: frequency = \:  \boxed{  ?}\:  \\ v_{w} =  14 50 \\ but \to \\ v = f \gamma   \to \:  \gamma  =  \frac{v}{f}  : hence \to \\  \gamma _{w} =  \frac{v_{w}}{f_{w}}   =  \frac{1}{100}  \times \gamma _{r}  =  \frac{1}{100}  \times 857,142.85714  \\\gamma _{w}  =  \boxed{8,571.4285714 \: m} : hence \to \:  \\ f_{w} =  \frac{v_{w}}{ \gamma _{w}}  =  \frac{1450}{8,571.4285714}  =  \boxed{0.1691666667} \\ if \: the \: number \: of \: times = \boxed{ x} \\ f_{r} (x)=f_{w} \\ (x) =  \frac{f_{w}}{f_{r}}  =  \frac{0.1691666667}{350}  = 0.0004833333 \\ hence \to \\ the  \: frequency  \: of \:  the \:  radio  \: wave  \: is \to \:   \boxed{{4.8 \times 10}^{ - 4}  }\:  \\ that  \: of  \: the \:  wave  \: created  \: in  \: the  \: water.

♨Rage♨

8 0
3 years ago
A particle of mass 4.5 × 10-8 kg and charge +5.4 μC is traveling due east. It enters perpendicularly a magnetic field whose magn
egoroff_w [7]

Answer:

0.00970 s

Explanation:

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

Force due to magnetic field = qvB sin θ

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v = velocity of the charge

B = magnetic field strength = 2.7 T

θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

where w = angular velocity.

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

mvw = qvB

mw = qB

w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)

w = 3.24 × 10² rad/s

w = 324 rad/s

w = (angular displacement)/time

Time = (angular displacement)/w

Angular displacement = π rads (half of a circle; 2π/2)

Time = (π/324) = 0.00970 s

Hope this Helps!!!

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Choose the correct answer from the given alternatives
Black_prince [1.1K]

Answer:

d

Explanation:

cggbhhgffffggdfvvcdfh

8 0
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