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3 years ago

Select the correct answer from each drop-down menu.

Physics

2 answers:

Zina [86]3 years ago

6 0

Answer:

I agree with the other guy, I think its lower to higher.

Explanation:

fgiga [73]3 years ago

5 0

Low place to a high place I think

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Alkaline batteries have the advantage of putting out constant voltage until very nearly the end of their life. How long will an

Sidana [21]

Answer:

Explanation:

Voltage, V = 1.58 V

Power, P = 1 W

1 A.h

Charge, Q = 1 A.h = 1 x 3600 A.s = 3600 C

Power x time = Voltage x charge

1 x t = 1.58 x 3600

t = 1.58 x 3600 second

t = 1.58 hours

8 0

3 years ago

A straight wire of length 4.5 cm moves at a constant speed of 5.2m/s perpendicular to its length and a uniform magnetic field. I

liraira [26]

Well you have to minus the 4.5 to 5.2 and the answer to that would be -11.5 and calculated that to be 4.5

3 0

3 years ago

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000 cycles

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2 years ago

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2 years ago