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3 years ago

7

The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car’s moti

on?

2 answers:

Nitella [24]3 years ago

3 0

The car is staying still. either getting petrol or stopped at a light .

Novosadov [1.4K]3 years ago

3 0

Answer:

The car is not moving

Explanation:

On a position-time graph, the gradient of the line represents the velocity. If the gradient is positive, then the car is speeding up, if it is negative, then the car is slowing down, if it is a horizontal line (where the gradient is zero) then the velocity is zero.

Hence, the horizontal line indicated that the car is not moving.

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35 POINTS! Say If You Drop A Ball from 100 Centimeters. When The ball Bounces, The Ball Does Not Bounce Back Up To 100 Centimete

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some ball when you bounce it it comes back up but according to gravity the energy goes away

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3 years ago

a concave mirror has a focal length of 18 cm. where will an image form if an object is placed 58 cm from the mirror

MAXImum [283]

Answer:

here

Explanation:

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3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

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3 years ago

Suppose you were asked to demonstrate electromagnetic induction. Which of the following situations will result in an electric cu

steposvetlana [31]

D. All of the above. When a wire loop is moved or rotated in a magnetic field, there is a change in magnetic flux which produces emf in wire loop and hence electric current is produced.

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3 years ago

Read 2 more answers

What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude

Bas_tet [7]

Answer:

The fraction of kinetic energy to the total energy is $\frac{K}{T}=\frac{8}{9}$ .

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

$K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)$

The total energy is

$T =0.5 mw^2A^2 ..... (2)$

Divide (1) by (2)

$\frac{K}{T}=\frac{8}{9}$

5 0

3 years ago