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Yuliya22 [10]
4 years ago
9

A particle moves so that its position (in meters) as a function of time (in seconds) is r = i ^ + 4t2 j ^ + tk ^. Write expressi

ons for (a) its velocity and (b) its acceleration as functions of time.
Physics
2 answers:
mario62 [17]4 years ago
6 0

Answer:

a.V=8tj+k

b.a=8j

Explanation:

Given:

Position r= i+4t^2j +tk

Nb r is position in metre and time in seconds

a.velocity is change in position/ change in time

v= ∆r/∆t =dr/dt

V=d ( i+ 4t^2j+tk)/dr

Differenting with respect to (t)

V=8tj+K

b.acceleration = change in velocity/change in time

a= ∆v/∆r =dv/dt

a=d (8tj+k)/dt

a= 8j

GREYUIT [131]4 years ago
3 0

Answer:

(a) velocity, v = 8t j + k

(b) acceleration, a = 8 j

Explanation:

The position of the particle as a function of time is given as;

r = i + 4t² j + t k         --------------------(i)

(a) To get the expression of its velocity, v, find the derivative of its position with respect to time by differentiating equation (i) with respect to t as follows;

v = dr / dt = 0 + 8t j + k

v = dr / dt = 8t j + k

v = 8t j + k          ----------------------(ii)

Therefore, the equation/expression for the particle's velocity (v) is

v = 8t j + k

(b) To get the expression of its acceleration, a, find the derivative of its velocity with respect to time by differentiating equation (ii) with respect to t as follows;

a = dv / dt = t j + 0

a = dv / dt = t j

a = 8 j

Therefore, the expression for the particle's acceleration, a, is a = 8 j

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Explanation:

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3 years ago
A spool whose inner core has a radius of 1.00 cm and whose end caps have a radius of 1.50 cm has a string tightly wound around t
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Answer:

v₁ = 37.5 cm / s

Explanation:

For this exercise we can use that angular and linear velocity are related

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in the case of the spool the angular velocity for the whole system is constant,

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         w = v₀ /r₀

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since the angular velocity is the same we set the two expressions equal

        \frac{v_o}{r_o} = \frac{v_1}{r_1}

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3 years ago
two electronics students are discussing electrical units. student A says that electrical power is measured in units coulombs, st
Savatey [412]
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Question 20
oksian1 [2.3K]

Answer:

The image distance is 17.56 cm

Explanation:

We have,

Height of light bulb is 3 cm.

The light bulb is placed at a distance of 50 cm. It means object distance is, u =-50 cm

Focal length of the lens, f = +13 cm

Let v is distance between image and the lens. Using lens formula :

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{13}+\dfrac{1}{(-50)}\\\\v=17.56\ cm

So, the image distance is 17.56 cm.

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3 years ago
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