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3 years ago

What is the weight of 5kgs of apples on Earth?

Physics

1 answer:

gregori [183]3 years ago

3 0

Answer:5kgs

Explanation:

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An overhead electric power line carries a maximum current of 125 A. What is the magnitude of the maximum magnetic field at a poi

BARSIC [14]

Answer:

B= 55.6×10^(-7) Tesla

Explanation:

B= μoI/(2πr)

B: magnetic field strength

μo: permeability of free space and is equal to 4π×10^(-7) T.m/A

r: distance from the wire

I : current in the wire

B= (4π×10^(-7)×125)/(2π×4.5)

B= 55.6×10^(-7) Tesla

3 0

4 years ago

You hold a bucket in one hand. In the bucket is a 500 g rock. You swing the bucket so the rock moves in a vertical circle 2.2 m

slavikrds [6]

Answer:v=3.28 m/s

Explanation:

Given

mass of rock $m=500 gm$

diameter of circle $d=2.2 m$

radius $r=\frac{2.2}{2}=1.1 m$

At highest Point

$mg+N=\frac{mv^2}{r}$

At highest Point N=0 because mass is just balanced by centripetal Force

thus $mg=\frac{mv^2}{r}$

$v=\sqrt{gr}$

$v=\sqrt{9.8\times 1.1}$

$v=\sqrt{10.78}$

$v=3.28 m/s$

6 0

3 years ago

The ossicles (the three tiny bones in the middle ear) are responsible for __________.

Feliz [49]

Answer:

D.amplifying sound vibrations from the eardrum

this is correct

3 0

3 years ago

Read 2 more answers

Which expression shows the amount of work a 6-kilogram mass is capable of performing if it is dropped from a height of 2 meters?

Ipatiy [6.2K]

The acceleration of gravity is 9.8 m/s^2

3 0

3 years ago

Read 2 more answers

"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

7 0

3 years ago