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2 years ago

How much work is done when a hoist lifts a 290-kg rock to a height of 7 m?

1 answer:

5 0

Given: Mass m = 290 Kg; Height h = 7 m

Required: Work = ?

Formula: Work = Force x distance but, Force F = mg

W= fd

W = mgh

W= (290 Kg)(9.8 m/s²)(7 m)

W = 19,894 Kg.m²/s²

W = 19,894 J

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If a typical smelt weighs 225 g, what is the total mass of pcbs in a smelt in the great lakes? If a typical smelt weighs 225 g,

photoshop1234 [79]

Answer;

= 2.34 x 10^-4 g pcbs

Explanation;

-A typical smelt has 1.04 ppm of PCB is great lakes.

Therefore;

The amount of PCB in a smelt in great leak can be calculated by multiplying the mass of smelt by the pcbs present in a smelt

= 225 g * ( 1.04/100000)

= 2.34 x10^-4 g pcbs

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2 years ago

List some ways houses would be built differently if gravity were much stronger or much weaker ?two

tia_tia [17]

Answer:

If gravity were stronger, houses and buildings would have to be more sturdy so they do not get crushed to the ground. If gravity were weaker, buildings and houses would have to be built stronger so they do not float away.

Hope this helps you!

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2 years ago

Read 2 more answers

QUESTION 5

Fiesta28 [93]

Answer:

D) sudden and irresistible on sets of sleep during normal waking periods

Explanation:

D because Narcolepsy (sudden and irresistible on sets of sleep during normal waking periods) is a part of insomnia

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2 years ago

Carlos has eight marbles that are all the same size. he knows that one marble has a mass of 10 grams. to find out the total mass

AURORKA [14]

Carlos lost his marbles

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2 years ago

Two 13 cm -long thin glass rods uniformly charged to +11nC are placed side by side, 4.0 cm apart. What are the electric field st

DedPeter [7]

Answer:

E1 = 10.15 * 10^4 N/C

E2 = 0

E3 = 10.15 *10^4 N/C

Explanation:

Given data:

Two 13 cm-long thin glass rods ( L ) = 0.13 m

charge (Q) = +11nC

distance between thin glass rods = 4 cm .

<u>Calculate the electric field strengths </u>

electric charge due to a single glass rod in the question ( E ) = $\frac{Q}{2\pi e_{0}rL }$

equation 1 can be used to determine E1, E2 and E3 because the points lie within the two rods hence the net electric field produced will be equal to the difference in electric fields produced

applying equation 1 to determine E1

E1 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.01} - \frac{1}{0.03} )$ ( distance from 1 rod is 0.01 m and from the other rod is 0.03 )

= $\frac{11*10^{-9} }{2*3.14*8.85*10^{-12}*0.13 } ( 66.67 )$

= 10.15 * 10^4 N/C

applying equation 1 to determine E2

E2 = $\frac{Q}{2\pi e_{0}rL }$ $( \frac{1}{0.02} - \frac{1}{0.02} )$

therefore E2 = 0

E1 = E3

hence E3 = 10.15*10^4 N/C

4 0

2 years ago