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3 years ago

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Let the acceleration of a particle be given by a(t)=t-1. the initial velocity is v(0)=10. find v(5). what is the displacement of

the particle in between t=0 and t=5

1 answer:

Natalka [10]3 years ago

3 0

Here, check this out.

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What are frames of reference

Gnoma [55]

Answer:

Reference frame, also called frame of reference, in dynamics, system of graduated lines symbolically attached to a body that serve to describe the position of points relative to the body.

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3 years ago

Read 2 more answers

A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is a

Licemer1 [7]

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

v = w r

a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

v₁ = w 1,2 (boy)

v₂ = w 1.8 (girl)

whereby

v₂> v₁

reviewing the claims we have

a₁ = α 1,2

a₂ = α 1.8

a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

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3 years ago

Airplane tickets are extremely cheap right now. You decide to take a trip to Paris for a quick weekend vacation. You hop on the

andreyandreev [35.5K]

Answer:

i don't exactly know but im srry for not giving the right answer

Explanation:

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3 years ago

a bird flies at a speed of 15 m/s for 10s, 20 m/s for 20s, and 30 m/s for 15s. what is the birds average speed

gtnhenbr [62]

Average speed = (total distance) / (total time)

First segment:

15 m/s for 10 sec; distance covered = (15 x 10) = 150 m

Second segment:

20 m/s for 20 sec; distance covered = (20 x 20) = 400 m

Third segment:

30 m/s for 15 sec; distrance covered = (30 x 15) = 450 m

Total time = (10s + 20s + 15s) = 45 sec

Total distance = (150m + 400m + 450m) = 1,000 m (1 km)

<em>Average speed</em> = (1,000 m) / (45 sec) = <em>22.22 m/s</em>

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4 years ago

Emily takes a trip, driving with a constant velocity of 87.5 km/h to the north except for a 22 min rest stop. If Emily's average

rjkz [21]

<h2>Answer:</h2>

2.63 hours

<h2>Explanation:</h2>

First we have to read carefully the problem, It start telling us that <u>Emily </u>was driving with average velocity of <em>87.5 km/h to the north</em>. Velocity is a physical variable that is definded in terms of :

Speed of the object (<em>87.5 km/h</em>)

Direction of the object (<em>north</em>)

Them the problem says that <u>Emily's trip has an average velocity of </u><u><em>76.8 km/h to the north </em></u>because she made a rest stop for 22 minutes<u><em>.</em></u>

Now you can ask yourself: If Emily was driving at <em>87.5 km/h, why the average speed of the trip was </em><u><em>76.8 km/h?</em></u>

we need to see how speed is defined:

$Speed=\frac{distance}{time} =\frac{x}{y}$

<em>***Note 1: </em>

Distance will be called as<em> "x" </em>and units will be kilometers(<em>km</em>)

Time will be called as <em>"y" </em>and units will be hours(<em>h</em>)

So, 87.5 km/h would be the average speed for a distance "x" and a time "y" without stops. But Emily made a stop that took 22 minutes. For the same distance Emily's trip took more time and time is in the denominator. If our numerator is constant and our denominator gets higher, the final result will be lower

Now, we have the follow expressions:

$87.5=\frac{x}{y}~~(1) \\76.8=\frac{x}{y+\frac{22}{60}}~~(2)$

***Note 2:

we need to convert 22 <em>minutes</em> to <em>hours</em>. 1 <em>hour</em>= 60 <em>minutes</em>, so we need to apply the follow covert factor:

$y(hours)=y(minutes)*\frac{1(hour)}{60(minutes)}$

The problem aks us about how long does the trip take, it means that we have to find y.

We have two variables (x and y), and we have two equationts. We know that x have the same value for both problems (Because both average speeds have the same distance), so we can solve both equations for x and made equal each other

$87.5*y=x~(3)\\76.8*(y+\frac{22}{60})=x~(4)$

$87.5*y=76.8*(y+\frac{22}{60})~(5)$

We have to expand (5) and then we have to solve for y

$87.5*y=76.8*y+76.8*\frac{22}{60}~~(6)\\87.5*y-76.8*y=76.8*\frac{22}{60}~~(7)\\10.7*y=76.8*\frac{22}{60}~~(8)\\y=\frac{76.8}{10.7}*\frac{22}{60}(hours)~~(9)\\y=2.63(hours)~~(10)$

3 0

3 years ago