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3 years ago

Please help on this one

Physics

1 answer:

Oksi-84 [34.3K]3 years ago

6 0

When a hypothesis is confirmed true by evidence from many experiments, it becomes widely accepted by scientists, and it gets a new name: 'Theory' .

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You move a 25 N object 5.0 meters. How much energy did you transformed?

lilavasa [31]

You move a 25 N object 5.0 meters. How much energy did you transformed?

25n x 5m = 125 J

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3 years ago

Heat always flows from a what place to a what place

Vera_Pavlovna [14]

Answer:

Heat is always the transfer of energy from an object at a higher temperature to an object at a lower temperature.

Explanation:

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3 years ago

Read 2 more answers

Which of these boxes will not accelerate? A. B. C. D.

serg [7]

Answer: The correct answer is Image B.

Explanation: For an object to accelerate, there should be unbalanced forces present. An object will move in the direction of net force.

Balanced forces are defined as the forces acting on the same object which are equal in magnitude but act in opposite direction. The net forces are 0.

Unbalanced forces are defined as the forces acting on the same object which are unequal in magnitude. The net force is non-zero.

For the given images:

Image A: This box will accelerate easily because the net force is non-zero and is moving in right direction.

Image B: This box will not accelerate because the net force is zero as all the forces are balancing one another. Hence, the object will stay at rest.

Image C: This box will accelerate easily because the net force is non-zero and is acting in between the normal and applied force.

Image D: This box will accelerate easily because the net force is non-zero and is moving in right direction.

Hence, the correct option is Image B.

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3 years ago

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Water flows over a section of Niagara Falls at the rate of 1.2×106 kg/s and falls 50.0 m. How much power is generated by the fal

quester [9]

Answer:

5.88×10⁸ W

Explanation:

Power = energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

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3 years ago

A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

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2 years ago