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Scorpion4ik [409]

3 years ago

10

A box of mass 3 kg is lifted 1.5 m onto a shelf. Calculate the change in its gravitational potential energy. The gravitational f

ield strength is 10 N/kg.

2 answers:

Dmitrij [34]3 years ago

8 0

Answer:

yesssir....yup...thanks

Explanation:

THANKSSSSSSSSSS for free points

andrey2020 [161]3 years ago

5 0

Answer:

Nvm..give me Brainliest..I need to rank up..thanks (:

Explanation:

I will help you with your questions..

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What is someone who leaves there country to settle in another?

Eddi Din [679]

Answer:

an immigrant

Explanation:

8 0

3 years ago

Read 2 more answers

At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl

Debora [2.8K]

Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

$s=ut+0.5at^2 \\$

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

$h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9$

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

$h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\$

Solving both equations

$600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\$

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

$h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\$

Passing of B occurs at 4108.31 height.

6 0

3 years ago

A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Natasha2012 [34]

Explanation:

There are three points in time we need to consider. At point 0, the mango begins to fall from the tree. At point 1, the mango reaches the top of the window. At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

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What is the most appropriate tool to use when collecting data from the investigation

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If it is just a reseach investigatory project, you could monitor it by using different variables (controlled), acurate measuring devices and camera for documentations

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A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph

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A vacuum has no air particles within it, it is completely empty.

Therefore, any of the above that requiring particles to travel will not be able to cross it and the observer will not experience it.

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