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3 years ago

How much work should be done to lift a 5kg brick to the height of 12m

Physics

2 answers:

Goshia [24]3 years ago

8 0

Answer: 588 joules

Explanation:

Work is done when force is applied on an object over a distance ( whether vertical or horizontal). It is measured in joules.

Thus, Workdone = Force X distance

- Vertical distance to be moved by the brick = 12 metres

- Mass of box = 5kg

- Acceleration due to gravity when box was lifted represented by g is a constant with value of 9.8m/s^2

Now, recall that Force = Mass x acceleration due to gravity

i.e Force = 5kg x 9.8m/s^2

Force = 49 Newton

So, Workdone = Force X Distance

Workdone = 49 Newton X 12 metres

Workdone = 588 joules

Thus, 588 joules of work was done.

zhenek [66]3 years ago

8 0

Answer:

588 j

Explanation:

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A horizontal applied force of magnitude 25 N acts on a block sliding on a horizontal surface. The force of friction between the

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' A ' and ' D ' are both correct statements.

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3 years ago

On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he

Pachacha [2.7K]

Answer:

d) g/2

Explanation:

We need to use one of Newton's equations of motion to find the position of the stone at any time t.

x(t) = x₀(t) + ut - ¹/₂at²

Where

x₀(t) = initial position of the stone.

x(t) - x₀(t) = distance traveled by the stone at any time.

u = initial velocity of the stone

a = acceleration of the stone

t = time taken

On both planets, before the stone was thrown by the astronaut, x = 0 and t = 0.

=> 0 = x₀(t)

=> x₀(t) = 0

On earth, when the stone returns into the hand of the astronaut at time T on earth, x = 0.

=> 0 = 0 + uT - ¹/₂gT² (a = g)

=> uT = ¹/₂gT²

=> g = 2u/T

On planet X, when the stone returns into the hand of the astronaut, time = 2T , x = 0.

=> 0 = 0 + u(2T) - ¹/₂a(2T)²

=> 2uT = 2aT²

=> a = u/T

By comparing we see that a = g/2.

5 0

3 years ago

What assumption can be made about the densities of two different spheres that have

Triss [41]

Answer:

Their densities are different as well

Explanation:

Density is worked out by dividing mass by volume. If the mass was the same then the densities would be as well. But it's not.

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8 0

3 years ago

A golf ball is dropped from rest from a height of 9.5m. It hits the pavement then bounces back up rising just 5.7 m before fal

Oksanka [162]

Answer: 3.4s

Explanation:

There are three stages in the motion of the ball, so you have to calculate the times for every stage.

1) Ball dropping from 9.5m: free fall

d = Vo + gt² / 2

Vo = 0 ⇒ d = gt² / 2 ⇒ t² = 2d / g = 2 × 9.5 m / 9.81 m/s² = 1.94 s²

⇒ t = √ (1.94 s²) = 1.39s

2) Ball rising 5.7m (vertical rise)

i) Determine the initial speed:

Vf² = Vo² - 2gd

Vf² = 0 ⇒ Vo² = 2gd = 2 × 9.81 m/s² × 5.7m = 111.8 m²/s²

⇒ Vo = 10.6 m/s

ii) time rising

Vf = Vo - gt

Vf = 0 ⇒ Vo = gt ⇒

t = Vo / g = 10.6 m/s / 9.81 m/s² = 1.08 s

3) Ball dropping from 5.7 m to 1.20m above the pavement (free fall)

i) d = 5.7m - 1.20m = 4.5m

ii) d = gt² / 2 ⇒ t² = 2d / g = 2 × 4.5 m / 9.81 m/s² = 0.92 s²

⇒ t = √ (0.92 s²) = 0.96s

4) Total time

t = 1.39s + 1.08s + 0.96s = 3.43s ≈ 3.4s

4 0

3 years ago

Read 2 more answers

Azurite is a mineral that contains 55.1% of copper. How many meter of copper wire with diameter of 0.0113 in can be produced fro

soldier1979 [14.2K]

Answer:

1402.73 m

Explanation:

Mass of Azurite=3.25 lb

Percent of copper in AZurite mineral=55.1%

Diameter of copper wire,d=0.0113 in

Radius of copper wire= $r=\frac{d}{2}=\frac{0.0113}{2}=0.00565 in=\frac{565}{100000}=\frac{565}{100}\times \frac{1}{1000}=5.65\times 10^{-3}in$

$\frac{1}{1000}=10^{-3}$

Density of copper= $\rho=8.96g/cm^3$

1 lb=454 g

3.25 lb= $3.25\times 454=1475.5 g$

Mass of Azurite= $1475.5 g$

Mass of copper= $\frac{55.1}{100}\times 1475.5=813 g$

Density= $\frac{Mass}{volume}$

Using the formula

$8.96=\frac{813}{volume\;of\;copper}$

Volume of copper wire= $\frac{813}{8.96}=90.7cm^3$

Radius of copper wire= $5.65\times 10^{-3}\times 2.54=14.35\times 10^{-3} cm$

1 in=2.54 cm

Volume of copper wire= $\pi r^2 h$

$\pi=3.14$

Using the formula

$90.7=3.14\times (14.35\times 10^{-3})^2\times h$

$h=\frac{90.7}{3.14\times (14.35\times 10^{-3})^2}$

$h=140273 cm$

1 m=100 cm

$h=\frac{140273}{100}=1402.73 m$

Hence, the length of copper wire required=1402.73 m

7 0

3 years ago