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3 years ago

When a car’s velocity is negative and its acceleration is negative, what is happening to the car’s motion?

Physics

1 answer:

lilavasa [31]3 years ago

5 0

If the velocity and acceleration are negative it means the car is not moving.

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Given the potential on a spherical shell of radius R to be V (R, θ) = V0 cos θ, calculate V (r, θ) both inside and outside the s

avanturin [10]

Answer:

not sure

Explanation:

5 0

3 years ago

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,

Art [367]

Answer:

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

Explanation:

given,

length of ladder = 10 ft

let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

$2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0$ ..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

$16\times 1.4 + 12\dfrac{dy}{dt} = 0$

$\dfrac{dy}{dt} = -\dfrac{5.6}{3}$

let the angle between the ladders be θ

$tan\theta = \dfrac{y}{x}$

y = xtan θ

$\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}$

$-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}$

$\dfrac{25}{2} \dfrac{d\theta}{dt} =\dfrac{-17.5}{6}$

$\dfrac{d\theta}{dt} =-0.233\ rad/s$

6 0

3 years ago

If wavelength and speed of a wave are 4 m and 332 m/s respectively, calculate its frequency

Furkat [3]

Explanation:

Given 

wavelength = 4 m

speed = 332 m/s

frequency = ?

We know we have the formula 

wavelength = speed / frequency 

4 = 332 / frequency 

frequency = 332/4

Therefore frequency is 83 Hertz .

4 0

2 years ago

Unpolarized light with intensity I0I0I_0 is incident on an ideal polarizing filter. The emerging light strikes a second ideal po

zvonat [6]

Answer:

$0.293I_0$

Explanation:

When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.

Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

$I_1=\frac{I_0}{2}$

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

$I_2=I_1 cos^2 \theta$

where

$\theta$ is the angle between the axes of the two polarizer

Here we have

$\theta=40^{\circ}$

So the intensity after the 2nd polarizer is

$I_2=I_1 (cos 40^{\circ})^2=0.587I_1$

And substituting the expression for I1, we find:

$I_2=0.587 (\frac{I_0}{2})=0.293I_0$

5 0

2 years ago

ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements.

vekshin1

Answer:

The answer is 500

Explanation:

6 0

2 years ago