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2 years ago

13

Zoe is setting up a track for a toy car. The track has a ramp that is 32° above horizontal. If Zoe wants the car to travel as a

projectile for 1.0 seconds, how fast does the toy car need to be moving as it leaves the ramp?
4.9 m/s
1.0 m/s
9.2 m/s
7.4 m/s m

1 answer:

jolli1 [7]2 years ago

8 0

Answer:

Explanation:

Not enough information.

IF we ASSUME she wants the car to be at LAUNCH LEVEL after 1 second of flight.

THEN

The highest point will have zero vertical velocity and will have taken ½ second to get there. This means that the initial vertical velocity was

v = gt

vy₀ = 9.8(0.5)

vy₀ = 4.9 m/s

vsinθ = vy₀

v = vy₀/sinθ

v = 4.9/sin32

v = 9.2466...

v = 9.2 m/s

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Explica de que tipo es cada oración según la actitud del hablante

Volgvan

Answer:

Según la actitud del hablante las oraciones se clasifican en enunciativas, interrogativas, etc. ... adverbios o expresiones que complementan a toda la oración (COr): ojalá, quizá.

Explanation:

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2 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

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2 years ago

A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

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3 years ago

Problems with solar energy include _____.

german

First choice: the inability of current technology to capture
large amounts of the Sun's energy

Well, it's true that large amounts of it get away ... our 'efficiency' at capturing it is still rather low. But the amount of free energy we're able to capture is still huge and significant, so this isn't really a major problem.

Second choice: the inability of current technology to store
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No. We're pretty good at building batteries to store small amounts, or raising water to store large amounts. Storage could be better and cheaper than it is, but we can store huge amounts of captured solar energy right now, so this isn't a major problem either.

Third choice: inconsistencies in the availability of the resource

I think this is it. If we come to depend on solar energy, then we're
expectedly out of luck at night, and we may unexpectedly be out
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Fourth choice: lack of demand for solar energy

If there is a lack of demand, it's purely a result of willful manipulation
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3 years ago

Read 2 more answers

If a reaction starts with 30 grams how many should it end with ?

zysi [14]

30grams

Explanation:

If a reaction starts with 30grams then the reaction should end with 30grams.

This in conformity with the law of conservation of mass.

The law states that "in an isolated system, mass is neither created nor destroyed during chemical transformation".
Mass is the quantity of matter contained in a substance.
In chemical reactions, the mass of reactants must always be the same with the mass of the product baring any loss.
In an isolated system, there is no exchange of energy and mass.
Chemical systems are usually treated as isolated systems in which mass is conserved.

Learn more:

Chemical laws brainly.com/question/5896850

#learnwithBrainly

4 0

3 years ago