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3 years ago

8

According to this graph, the acceleration

1 answer:

skelet666 [1.2K]3 years ago

7 0

Answer:

1 m/s^2

Explanation:

a=(v-u)/t

where,

a=acceleration

v=final velocity

u=initial velocity

(4.5-0)/4.5

1 m/s^2

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Newton first law of motion

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3 years ago

A 1.1 kg block is initially at rest on a horizontal frictionless surface when a horizontal force in the positive direction of an

elixir [45]

Answer with Explanation:

Mass of block=1.1 kg

Th force applied on block is given by

F(x)= $(2.4-x^2)\hat{i}N$

Initial position of the block=x=0

Initial velocity of block= $v_i=0$

a.We have to find the kinetic energy of the block when it passes through x=2.0 m.

Initial kinetic energy= $K_i=\frac{1}{2}mv^2_i=\frac{1}{2}(1.1)(0)=0$

Work energy theorem:

$K_f-K_i=W$

Where $K_f=$ Final kinetic energy

$K_i$ =Initial kinetic energy

$W=Total work done$

Substitute the values then we get

$K_f-0=\int_{0}^{2}F(x)dx$

Because work done= $Force\times displacement$

$K_f=\int_{0}^{2}(2.4-x^2)dx$

$K_f=[2.4x-\frac{x^3}{3}]^{2}_{0}$

$K_f=2.4(2)-\frac{8}{3}=2.13 J$

Hence, the kinetic energy of the block as it passes thorough x=2 m=2.13 J

b.Kinetic energy = $K=2.4x-\frac{x^3}{3}$

When the kinetic energy is maximum then $\frac{dK}{dx}=0$

$\frac{d(2.4x-\frac{x^3}{3})}{dx}=0$

$2.4-x^2=0$

$x^2=2.4$

$x=\pm\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2x$

Substitute x= $\sqrt{2.4}$

$\frac{d^2K}{dx^2}=-2\sqrt{2.4}$

Substitute x= $-\sqrt{2.4}$

$\frac{d^2K}{dx^2}=2\sqrt{2.4}>0$

Hence, the kinetic energy is maximum at x= $\sqrt{2.4}$

Again by work energy theorem , the maximum kinetic energy of the block between x=0 and x=2.0 m is given by

$K_f-0=\int_{0}^{\sqrt{2.4}}(2.4-x^2)dx$

$k_f=[2.4x-\frac{x^3}{3}]^{\sqrt{2.4}}_{0}$

$K_f=2.4(\sqrt{2.4})-\frac{(\sqrt{2.4})^3}{3}=2.48 J$

Hence, the maximum energy of the block between x=0 and x=2 m=2.48 J

3 0

4 years ago

A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and

trapecia [35]

Answer:

a) $f'=544.66 \textup{Hz}$

b) $f'=584.75 \textup{Hz}$

Explanation:

Given:

Frequency of the whistle, f = 564 Hz

Radius of the circle, r = 71.2 cm = 0.712 m

Angular speed, ω = 17.1 rad/s

speed of source, $v_s$ = rω = 0.712 × 17.1 = 12.1752 m/s

speed of sound, v = 343 m/s

Now, applying the Doppler's effect formula, we have

$f'=f\frac{v\pm v_d}{v\pm v_s}$

where,

$v_d$ = relative speed of the detector with respect to medium = 0

a) for lowest frequency, we have the formula as:

$f'=f\frac{v}{v+v_s}$

on substituting the values, we get

$f'=564\times\frac{343}{343+12.1752}$

or

$f'=544.66 \textup{Hz}$

b) for maximum frequency, we have the formula as:

$f'=f\frac{v}{v-v_s}$

on substituting the values, we get

$f'=564\times\frac{343}{343-12.1752}$

or

$f'=584.75 \textup{Hz}$

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3 years ago