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2 years ago

One half of a balanced chemical equation is shown.

Physics

2 answers:

Zarrin [17]2 years ago

6 0

Answer:

A: 3 Mg, 2 P, 14 O, 12 H

Explanation:

I took the test on Edgenuity

Tpy6a [65]2 years ago

6 0

Answer:

A: 3 Mg, 2 P, 14 O, 12 H

Explanation:

i took the test responsibily

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A scientist is trying to predict the regions of the world most likely to see a decrease in natural resources from population cha

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Developed countries will see a decrease in natural resources, because their population will decrease.

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If the Big Bang theory is correct, then the universe is billions of years old. And if the Big Bang theory is correct, then the u

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Answer:

Deductive

Explanation:

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The question bases the conclusion on the premise that the Big Bang theory which is based on scientific evidence is several billions of years old and if the theory is correct the universe (as it is today) was not created six days

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3 years ago

What is the minimum v (in km/s) the rocket engines must provide to allow the craft to escape from the Earth?

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11 kilometers (7 miles) per second, or over 40,000 kilometers per hour (25,000 miles per hour)

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3 years ago

What is the power of a student that has done a work of 10 joules in 10 seconds

Alla [95]

Answer:

1 Watt

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3 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

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3 years ago