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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?

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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An isolated conducting sphere has a 12 cm radius. One wire carries a current of 1.0000049 A into it while another wire carries a

tatuchka [14]

Answer:

it will take for the sphere to increase in potential by 1500 V, 503.71 s.

Explanation:

The charge on the sphere after t seconds is:

q = (1.0000049 - 1.0000000) t = 0.0000049 t

The voltage on the surface is

V = k * $\frac{q}{R}$ = k 0.0000049 t / R

solve for t

t = (R*V) / (0.0000049 k) = (0.12 * 1500) / (0.0000049 * $9e^{9}$ ) = 503.71 s

7 0

4 years ago

To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T

Fantom [35]

Answer:

$V_1= 3.4*10^7m/s$

Explanation:

From the question we are told that

Nucleus diameter $d=5.50-fm$

a 12C nucleus

Required kinetic energy $K=2.30 MeV$

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

$K_2 +U_2=K_1+U_1$

where

$K_1$ =initial kinetic energy

$K_2$ =final kinetic energy

$U_1$ =initial electric potential

$U_2$ =final electric potential

mathematically

$U_2 = \frac{Kq_pq_c}{r_2}$

where

$r_f$ =distance b/w charges

$q_c$ =nucleus charge $=6(1.6*10^-^1^9C)$

$K$ =constant

$q_p$ =proton charge

Generally kinetic energy is know as

$K=\frac{1}{2} mv^2$

Therefore

$U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1$

Generally equation for radius is $d/2$

Mathematically solving for radius of nucleus

$R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})$

$R=2.75*10^-^1^5m$

Generally we can easily solving mathematically substitute into v_1

$q_p$ $=6(1.6*10^-^1^9C)$

$K_1=9.0*10^9 N-m^2/C^2$

$U_1= 0$

$R=2.75*10^-^1^5m$

$K=2.30 MeV$

$m= 1.67*10^-^2^7kg$

$V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }$

$V_1= 3.4*10^7m/s$

Therefore the proton must be fired out with a speed of $V_1= 3.4*10^7m/s$

8 0

3 years ago

Given a force of 88 N and an acceleration of 4 m/s2, what is the mass?

WARRIOR [948]

Newton's second law of motion says:

Force = (mass) x (acceleration)

88 = (mass) x (4)

Divide each side by 4 :

Mass = 22 kg.

7 0

4 years ago

A car travels 40 kilometers at an average speed of 80 km/h and then travels 40 kilometers at

miv72 [106K]

Answer:53.3m/s

Explanation:

time for first travel

time=distance/speed

time=40/80

time=0.5seconds

time for second travel:

time=distance/speed

time=40/40

time=1seconds

total time=0.5+1

total time=1.5seconds

Then average speed for 80km trip:

Speed=distance/time

Speed=80/1.5

Speed=53.3m/s

3 0

4 years ago

Wilma I. Ball walks at a constant speed of 5.93 m/s along a straight line from point A to point B and then back from B to A at a

rosijanka [135]

Answer:

Case:1 a) average speed $v_{avg}=4.1144\ m.s^{-1}$

Since the body returns to its initial position hence the displacement is zero so velocity is also zero.

CASE: 2 a) average speed $v_{avg}=3.7093\ m.s^{-1}$

Since the body returns to its initial position hence the displacement is zero so velocity is also zero.

Explanation:

Given:

Case: 1

speed of Wilma from point A to B, $v_{ab}=5.93\ m.s^{-1}$

speed of Wilma from point B to A, $v_{ba}=3.15\ m.s^{-1}$

Let the displacement from A to B be x meter.

Then the total displacement = 2x meter

time taken in going from A to B:

$t_{ab}=\frac{x}{5.93}\ s$

time taken in going from B to A:

$t_{ba}=\frac{x}{3.15} \ s$

<u>Wilma's average speed over the entire trip:</u>

$v_{avg}=\frac{2x}{t_{ab}+t_{ba}}$

$v_{avg}=2x\div (\frac{x}{5.93}+\frac{x}{3.15})$

$v_{avg}=4.1144\ m.s^{-1}$

Since the body returns to its initial position hence the displacement is zero so velocity is also zero.

CASE: 2

speed of Wilma from point A to B, $v_{ab}=4.51\ m.s^{-1}$

speed of Wilma from point B to A, $v_{ba}=3.15\ m.s^{-1}$

Let the displacement from A to B be x meter.

Then the total displacement = 2x meter

time taken in going from A to B:

$t_{ab}=\frac{x}{4.51}\ s$

time taken in going from B to A:

$t_{ba}=\frac{x}{3.15} \ s$

<u>Wilma's average speed over the entire trip:</u>

$v_{avg}=\frac{2x}{t_{ab}+t_{ba}}$

$v_{avg}=2x\div (\frac{x}{4.51}+\frac{x}{3.15})$

$v_{avg}=3.7093\ m.s^{-1}$

Since the body returns to its initial position hence the displacement is zero so velocity is also zero.

3 0

3 years ago