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4 years ago

The point upon which a lever turns (rotates) is called a

Physics

2 answers:

Pavlova-9 [17]4 years ago

8 0

Answer:fulcrum

Explanation: there is none

Ede4ka [16]4 years ago

3 0

The point upon which a lever turns is called a B) fulcrum.

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A scientific theory _______.

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B. is not a validated bu experimentation

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3 years ago

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Which action do researchers take to make advances in science?

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<span>b. They share their experimental results with other researchers</span>

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A number of conditions are required for a population to be in Hardy-Weinberg equilibrium. Which of the following are correct des

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3 years ago

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2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

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3 years ago

The dimensions of aluminum foil in a box for sale in super markets are 66 2/3 yards by 12 inches. the mass of the foil is 0.83 k

disa [49]

Length of the sheet is given as

$L = \frac{200}{3} yards = 6096 cm$

width of the sheet is given as

$w = 12 inches = 30.48 cm$

now let say its thickness is "t"

so the volume of the sheet is given as

$V = L*w*t$

$V = 6096*30.48* t$

$V = 185806.08*t cm^3$

mass of the sheet is given as

$m = 0.83 kg = 830 gram$

now we have

$density = \frac{mass}{volume}$

$2.70 = \frac{830}{185806.08*t}$

by solving above we have

$t = 1.65 * 10^{-3} cm$

so the thickness of sheet will be above

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3 years ago