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valentinak56 [21]

4 years ago

10

Suppose that you lift four boxes individually, each at a constant velocity. The boxes have weights of 3.0 N, 4.0 N, 6.0 N, and 2

.0 N, and you do 12 J of work on each. Match each box to the vertical distance through which it is lifted.

1 answer:

Alona [7]4 years ago

8 0

Answer:

The vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

Explanation:

Worked : work can be defined as the product of force and distance.

The S.I unit of work is Joules (J).

Mathematically it can be represented as,

W = F×d.................. Equation 1

d = W/F.............................. Equation 2

where W = work, F = force, d = distance.

<em>Given: W = 12 J</em>

(i) for the 3.0 N weight,

using equation 2

d = 12/3

d= 4 m.

(ii) for the 4.0 N weight,

d = 12/4

d = 3 m.

(iii) for the 6.0 N weight,

d = 12/6

d = 2 m.

(iv) for the 2.0 N weight,

d = 12/2

d = 6 m

Therefore vertical distance of weight 3.0 N = 4 m, vertical distance of weight 4.0 N = 3 m, vertical distance of weight 6.0 N = 2 m, vertical distance of weight 2.0 N = 6 m

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mariarad [96]

Answer:

$g_{moon}=1.67 [m/s^{2} ]$

Explanation:

The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.

$w =m*g\\$

where:

w = weight [N]

m = mass = 0.06 [kg]

g = gravity acceleration = 10 [N/kg]

Therefore:

$w=0.06*10\\w=0.6[N]$

By Hooke's law we know that the force in a spring can be calculated by means of the following expression.

$F=W\\F = k*x$

where:

k = spring constant [N/m]

x = deformed distance = 6 [cm] = 0.06 [m]

We can find the spring constant.

$k= F/x\\k=0.6/0.06\\k=10 [N/m]$

Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.

$F_{moon}=k*0.01\\F = 10*0.01\\F=0.1[N]$

Since this force is equal to the weight, we can now determine the gravitational acceleration.

$F=m*g_{moon}\\g=F/m\\g = 0.1/0.06\\g_{moon} = 1.67[m/s^{2} ]$

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3 years ago

Match each literary device with the excerpt in which it is used.

kirill [66]

Answer:

euphemism oxymoron dramatic irony Excerpts Literary Devices In this story I read, Karen assumes her sister Mira is flushed with joy at learning of her eldest son's imminent return from the battle front. But the narrator makes it clear that Mira is beginning to wonder about the anti-war letters she recently sent to the president. "Marla, don't tease Sandy about how much she's been eating lately. For all you know she might have another bun in the oven." Unsure if he was really safe from the tiger he spotted earlier, Johan's heart beat quickly as he crouched and listened to the screaming silence of the forest. I rights reserved

3 0

3 years ago

Read 2 more answers

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

A cheerleader lifts his 59.6 kg partner straight up off the ground a distance of 0.749 m before

Airida [17]

m = mass of the partner which the cheerleader lifts = 59.6 kg

h = height to which the partner is lifted by the cheerleader = 0.749 m

g = acceleration due to gravity = 9.8 m/s²

work done by the cheerleader in lifting the partner is same as the potential energy gained by the partner.

W = work done by the cheerleader in lifting the partner

PE = potential energy gained

so W = PE

potential energy is given as

PE = mgh

hence

W = mgh

inserting the values in the above formula

W = 59.6 x 9.8 x 0.749

W = 437.5 J

this is the work done in lifting the partner once.

the cheerleader does this 30 times , hence the total work done is given as

W' = 30 W

W' = 30 x 437.5

W' = 13125 J

5 0

4 years ago

What is the difference between mutual flux, leakage flux and magnetizing flux<br>

Lemur [1.5K]

In simple words, flux can be stated as the rate of flow of a fluid, radiant energy, or particles across a given area.

<u>Explanation:</u>

<u>Mutual Flux:</u>

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The SI unit of mutual flux is the Henry

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The SI unit of magnetic flux is the Weber

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3 years ago