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Trava [24]
3 years ago
14

If a person is driving a speed of 162 km and drive for 35 seconds. How far will they drive?

Physics
1 answer:
patriot [66]3 years ago
7 0

the correct answer is 162 km

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Electrostatic forces between charges depend on the product of
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We should also mention the item about whether the charges are
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What is the apparent magnitude of the brightest star in the Big Dipper?
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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
IgorLugansk [536]

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8 0
3 years ago
Consult interactive solution 2.22 before beginning this problem. a car is traveling along a straight road at a velocity of +30.0
Inessa05 [86]

Let a_1 be the average acceleration over the first 2.46 seconds, and a_2 the average acceleration over the next 6.79 seconds.

At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let v be the velocity of the car after the first 2.46 seconds.

By definition of average acceleration, we have

a_1=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

a_2=\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}

and we're also told that

\dfrac{a_1}{a_2}=1.66

(or possibly the other way around; I'll consider that case later). We can solve for a_1 in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:

1.66a_2=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

\implies1.66\left(\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}\right)=\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}

Now we can solve for v. We find that

v=20.8\,\dfrac{\mathrm m}{\mathrm s}

In the case that the ratio of accelerations is actually

\dfrac{a_2}{a_1}=1.66

we would instead have

\dfrac{15.2\,\frac{\mathrm m}{\mathrm s}-v}{6.79\,\mathrm s}=1.66\left(\dfrac{v-30.0\,\frac{\mathrm m}{\mathrm s}}{2.46\,\mathrm s}\right)

in which case we would get a velocity of

v=24.4\,\dfrac{\mathrm m}{\mathrm s}

6 0
3 years ago
Find an equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4.
m_a_m_a [10]

An equation in x and y for the line tangent to the curve ()=4,()=cos() at the point where =4 is x(t)=2t+2,y(t)=t^4.

<h3>What is tangent?</h3>

In calculation, the digression line to a plane bend at a given point is the straight line that "simply contacts" the bend by then. Leibniz characterized it as the line through a couple of boundlessly close focuses on the bend. The chart of digression is intermittent, implying that it rehashes the same thing endlessly. In contrast to sine and cosine in any case, digression has asymptotes isolating every one of its periods. The space of the digression capability is all genuine numbers with the exception of at whatever point cos⁡(θ)=0, where the digression capability is vague. Assuming they stroll in an orderly fashion, they are fundamentally following a digression way for the shape that is made inside the fencing.

Learn more about tangent, refer:

brainly.com/question/12585907

#SPJ4

7 0
2 years ago
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