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GrogVix [38]
3 years ago
7

A point moves on the x-axis in such a way that its velocity at time t (t > 0) is given by v=ln t/t . At what value of t does

v attain its maximum?
Physics
1 answer:
Olenka [21]3 years ago
5 0

Answer:

Explanation:

Given

Velocity of point is given by v=\frac{\ln t}{t}

To get maximum or minimum velocity differentiate v w.r.t t

\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{\frac{1}{t}\times t-1\times \ln(t)}{t^2}

so 1-\ln (t) should be equal to zero

\ln (t)=1

t=e

i.e. t=2.718\ s

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Light waves can be easily blocked but ______ waves pass through all substances? ( fill in the blank)
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Starting from rest, a basketball rolls from top of a hill to the bottom, reaching a translational speed of 6.8 m/s. Ignore frict
kkurt [141]

Answer:

Explanation:

for baseball

(a) Let the mass of the baseball is m.

radius of baseball is r.

Total kinetic energy of the baseball, T = rotational kinetic energy + translational kinetic energy

T = 0.5 Iω² + 0.5 mv²

Where, I be the moment of inertia and ω be the angular speed.

ω = v/r

T = 0.5 x 2/3 mr² x v²/r² + 0.5 mv²

T = 0.83 mv²

According to the conservation of energy, the total kinetic energy at the bottom is equal to the total potential energy at the top.

m g h = 0.83 mv²

where, h be the height of the top of the hill.

9.8 x h = 0.83 x 6.8 x 6.8

h = 3.93 m

(b) Let the velocity of juice can is v'.

moment of inertia of the juice can = 1/2mr²

So, total kinetic energy

T = 0.5 x I x ω² + 0.5 mv²

T = 0.5 x 0.5 x m x r² x v²/r² + 0.5 mv²

m g h = 0.75 mv²

9.8 x 3.93 = 0.75 v²

v = 7.2 m/s

7 0
3 years ago
At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet
expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

also the PE is given as function of position as

PE = \frac{1}{2}m\omega^2x^2

now it is given that

KE = PE

now we will have

\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2

A^2 - x^2 = x^2

2x^2 = A^2

x = \frac{A}{\sqrt2}

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

KE = \frac{1}{2}m\omega^2(A^2 - x^2)

now to find the fraction of kinetic energy

f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}

f = \frac{A^2 - (\frac{A}{3})^2}{A^2}

f_k = \frac{8}{9}

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

f_{PE} = 1 - f_k

f_{PE} = 1 - (8/9) = 1/9

7 0
3 years ago
Under which condition does Ohm's law apply? a. The current must be constant b. The power must be constant c. The temperature mus
ruslelena [56]

Answer:

option c

Explanation:

The temperature must be constant. Ohms law states that the current running through a conductor is directly proportional to the potential difference across it provided the temperature remains constant

6 0
2 years ago
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