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2 years ago

If the velocity of a bycycle is 10 m/s, how long does it take to cover a distance of 18 km? plzzz help me

Physics

2 answers:

insens350 [35]2 years ago

8 0

Answer:

1800 seconds

Explanation:

Given,

Distance ( d ) = 18 km

Velocity ( v ) = 10 m/s

To find : Time ( t ) = ?

Distance in km need to be converted to m.

1 km = 1000 m

18 km

= 18 x 1000

= 18000 m

So,

d = 18000 m

Formula : -

v = d / t

t = d / v

= 18000 / 10

t = 1800 seconds.

Therefore,

1800 seconds long, it will take to cover a distance of 18 km.

pashok25 [27]2 years ago

4 0

Your answer would 1800 seconds.

You're welcome ;)

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Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degree

kvv77 [185]

Answer:

When second polarizer is removed the intensity after it passes through the stack is

$I_f_3 = 27.57 W/cm^2$

2 When third polarizer is removed the intensity after it passes through the stack is

$I_f_2 = 102.24 W/cm^2$

Explanation:

From the question we are told that

The angle of the second polarizing to the first is $\theta_2 = 21^o$

The angle of the third polarizing to the first is $\theta_3 = 61^o$

The unpolarized light after it pass through the polarizing stack $I_u = 60 W/cm^2$

Let the initial intensity of the beam of light before polarization be $I_p$

Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

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

Now according to Malus’ law the intensity of light that would emerge from the second polarizing filter is mathematically represented as

$I_2 = I_1 cos^2 \theta_1$

$= \frac{I_p}{2} cos ^2 \theta_1$

The intensity of light that will emerge from the third filter is mathematically represented as

$I_3 = I_2 cos^2(\theta_2 - \theta_1 )$

$I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]$

making $I_p$ the subject of the formula

$I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}$

Note that $I_u = I_3$ as $I_3$ is the last emerging intensity of light after it has pass through the polarizing stack

Substituting values

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}$

$I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}$

$=234.622W/cm^2$

When the second is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

$I_f_3 = \frac{I_p}{2} cos ^2 \theta_2$

$I_f_3$ is the intensity of the light emerging from the stack

substituting values

$I_f_3 = \frac{234.622}{2} * cos^2(61)$

$I_f_3 = 27.57 W/cm^2$

When the third polarizer is removed the second polarizer becomes the

the final polarizer and the intensity of light emerging from the stack would be

$I_f_2 = \frac{I_p}{2} cos ^2 \theta_1$

$I_f_2$ is the intensity of the light emerging from the stack

Substituting values

$I_f_2 = \frac{234.622}{2} cos^2 (21)$

$I_f_2 = 102.24 W/cm^2$

7 0

2 years ago

A ball is thrown horizontally from the top of a building at 2 m/s. It takes 3 seconds to reach the ground. How far did the ball

svlad2 [7]

Every second it travels 2 meters and it traveled 3 so (2 x 3) would be 6meters it traveled

4 0

3 years ago

Click the Run Now button to start the simulations. Select "Many rays" and click the Screen checkbox. You should see a lamp and a

Gnoma [55]

Answer:

Answer explained below

Explanation:

(a) The rays are diverging near the lens. They change the direction when they passed through the converging lens

(b) If the light rays don't bend they will move away from the optical (principal axis) as the other waves are moving.

(c) If we decrease the distance between lens and light source, most of the rays diverge and no ray converges on the screen even after passing through the lens. Here is a screenshot.

5 0

3 years ago

A 1.0-μm-diameter oil droplet (density 900 kg/m3) is negatively charged with the addition of 39 extra electrons. It is released

GREYUIT [131]

Answer:

$6.75\mu C/m^2$

Explanation:

We are given that

Diameter,d= $1\mu m=1\time 10^{-6} m$

$1\mu m=10^{-6} m$

Radius,r= $\frac{d}{2}=\frac{1}{2}\times 10^{-6}=0.5 \times 10^{-6} m$

Density, $\rho=900kg/m^3$

Total number of electrons,n=39

Charge on electron = $1.6\times 10^{-19} C$

Total charge= $q=ne=39\times 1.6\times 10^{-19}=62.4\times 10^{-19} C$

Distance,s=2mm= $2\times 10^{-3} m$

Mass = $density\times volume=900\times \frac{4}{3}\pi r^3=900\times \frac{4}{3}\pi(0.5\times 10^{-6})^3=4.7\times 10^{-16} kg$

Initial velocity,u=0

Final speed,v=4.5 m/s

$v^2-u^2=2as$

$(4.5)^2-0=2a(2\times 10^{-3})$

$20.25=4a\times 10^{-3}$

$a=\frac{20.25}{4\times 10^{-3}}=5062.5m/s^2$

Force,F=ma

$qE=ma$

$q(\frac{\sigma}{2\epsilon_0})=ma$

$\sigma=\frac{2\epsilon_0ma}{q}=\frac{2\times 8.85\times 10^{-12}\times 4.7\times 10^{-16}\times 5062.5}{62.4\times 10^{-19}}$

$\epsilon_0=8.85\times 10^{-12}$

$\sigma=6.75\times 10^{-6}C/m^2=6.75\mu C/m^2$

6 0

3 years ago

On a frictionless surface how much force is necessary to accelerate a 0.49 kg object to the left at 4.8 m/s2?

Alenkasestr [34]

Answer:

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

force = 0.49 × 4.8 = 2.352

We have the final answer as

Hope this helps you

8 0

2 years ago

If the velocity of a bycycle is 10 m/s, how long does it take to cover a distance of 18 km? plzzz help me​

If the velocity of a bycycle is 10 m/s, how long does it take to cover a distance of 18 km? plzzz help me