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3 years ago

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The potential difference between the plates of an air-filled parallel-plate capacitor with a plate separation of 4 cm is 51 V. W

hat is the magnitude of the electric field between the plates of this capacitor

1 answer:

Salsk061 [2.6K]3 years ago

6 0

Answer:

Explanation:

Electric field between plates of a parallel plate capacitor is uniform .

In a uniform electric field , relation between electric field and potential gradient is as follows

electric field = potential gradient [ E = - dV / dl ]

in the given case ,

dV = 51 V ,

dl = 4 cm

= 4 x 10⁻² m

E = 51 / 4 x 10⁻²

= 12.75 x 10² V / m

= 1275 V / m

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A ball is thrown horizontally from the top of a 60 m building and lands 100 m from the base of the building. How long is the bal

zhannawk [14.2K]

Answer:

The ball is in the air for 3.5 seconds

The initial horizontal component of velocity is 28.6 m/s

The vertical component of the final velocity is 34.3 m/s downward

The final velocity is 44.7 m/s in the direction 50.2° below the horizontal

Explanation:

A ball is thrown horizontally

That means the vertical component of the initial velocity $u_{y}=0$

The initial velocity is the horizontal component $u_{x}$

The ball is thrown from the top of a 60 m

That means the vertical displacement component y = 60 m

→ y = $u_{y}$ t + $\frac{1}{2}$ gt²

where g is the acceleration of gravity and t is the time

y = -60 m , g = -9.8 m/s² , $u_{y}=0$

Substitute these values in the rule

→ -60 = 0 + $\frac{1}{2}$ (-9.8)t²

→ -60 = -4.9t²

Divide both sides by -4.9

→ 12.2449 = t²

Take √ for both sides

∴ t = 3.5 seconds

* <em>The ball is in the air for 3.5 seconds </em>

The initial velocity is the horizontal component $u_{x}$

The ball lands 100 meter from the base of the building

That means the horizontal displacement x = 100 m

→ x = $u_{x}$ t

→ t = 3.5 s , x = 100 m

Substitute these values in the rule

→ 100 = $u_{x}$ (3.5)

Divide both sides by 3.5

→ $u_{x}$ = 28.57 m/s

<em>The initial horizontal component of velocity is 28.6 m/s</em>

The vertical component of the final velocity is $v_{y}$

→ $v_{y}$ = $u_{y}$ + gt

→ $u_{y}$ = 0 , g = -9.8 m/s² , t = 3.5 s

Substitute these values in the rule

→ $v_{y}$ = 0 + (-9.8)(3.5)

→ $v_{y}$ = -34.3 m/s

<em>The vertical component of the final velocity is 34.3 m/s downward</em>

The final velocity v is the resultant vector of $v_{x}$ and $v_{y}$

→ Its magnetude is $v=\sqrt{(v_{x})^{2}+(v_{y})^{2}}$

→ Its direction $tan^{-1}\frac{v_{y}}{v_{x}}$

→ $v_{y}$ = 28.6 , $v_{y}$ = -34.3

Substitute this values in the rules above

→ $v=\sqrt{(28.6)^{2}+(-34.3)^{2}}=44.66$

→ Its direction $tan^{-1}\frac{-34.3}{28.6}=-50.18$

The negative sign means the direction is below the horizontal

<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>

7 0

2 years ago

Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam

Hatshy [7]

Think of it like a graph. You start at the origin which is (0,0). go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6). 1 to the east, ((4,6). Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite.

Hope this helps!

6 0

2 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

3 years ago

A train is moving in a straight railway where it covered one third of the distance with

Alexxandr [17]

Answer:

<em>The average speed of the train is 45 km/h</em>

Explanation:

<u>Speed</u>

It's defined as the distance (d) per unit of time (t) traveled by an object. The formula is:

$\displaystyle v=\frac{d}{t}$

Let's call x the total distance covered by the train. It covered d1=1/3x with a speed of v1=25 km/h. The time taken is calculated solving for t:

$\displaystyle t_1=\frac{d_1}{v_1}$

$\displaystyle t_1=\frac{1/3x}{25}$

$\displaystyle t_1=\frac{x}{75}$

Now the rest of the distance:

d2 = x - 1/3x = 2/3x

Was covered at v2=75 km/h. Thus the time taken is:

$\displaystyle t_2=\frac{d_2}{v_2}$

$\displaystyle t_2=\frac{2/3x}{75}$

$\displaystyle t_2=\frac{2x}{225}$

The total time is:

$\displaystyle t_t=\frac{x}{75}+\frac{2x}{225}$

$\displaystyle t_t=\frac{3x}{225}+\frac{2x}{225}$

$\displaystyle t_t=\frac{5x}{225}$

Simplifying:

$\displaystyle t_t=\frac{x}{45}$

The average speed is the total distance divided by the total time:

$\displaystyle \bar v=\frac{x}{\frac{x}{45}}$

Simplifying:

$\boxed{\displaystyle \bar v=45\ km/h}$

The average speed of the train is 45 km/h

5 0

3 years ago

The acceleration due to gravity on the surface of Mars is about one-third the acceleration due to gravity on Earthâ€™s surface.

aksik [14]

Answer:

one-third of its weight on Earth's surface

Explanation:

Weight of an object is = W = m*g

Gravity on Earth = g₁ = 9.8 m/s

Gravity on Mars = g₂ = $\frac{1}{3}$ g₁

Weight of probe on earth = w₁ = m * g₁

Weight of probe on Mars = w₂ = m * g₂ -------- ( 1 )

As g₂ = g₁/3 --------- ( 2 )

Put equation (2) in equation (1)

so

Weight of probe on Mars = w₂ = m * g₁ /3

Weight of probe on Mars = $\frac{1}{3}$ m * g₁ = $\frac{1}{3}$ w₁

⇒Weight of probe on Mars = $\frac{1}{3}$ Weight of probe on earth

6 0

3 years ago