A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, its water level rises to 20.3 ml.

Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m

alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = $0.407\times 10^{-3}\ m$

Energy stored in the capacitor (U) = $7.87\ nJ=7.87\times 10^{-9}\ J$

Assuming the medium to be air.

So, permittivity of space (ε) = $8.854\times 10^{-12}\ F/m$

Area of the square plates is given as:

$A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2$

Capacitance of the capacitor is given as:

$C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F$

Now, we know that, the energy stored in a parallel plate capacitor is given as:

$U=\frac{CE^2d^2}{2}$

Rewriting in terms of 'E', we get:

$E=\sqrt{\frac{2U}{Cd^2}}$

Now, plug in the given values and solve for 'E'. This gives,

$E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C$

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0

3 years ago

A piano tuner is using a 392 Hz tuning fork to tune the wire for a G-Natural note. She hears 4 beats per second. What are the tw

inysia [295]

A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,

$f_{beat} =| f_2\pm f_1 |$

The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be

$f_{beat} =|392+4|= 396Hz$

$f_{beat} =|392-4|=388Hz$

Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz

5 0

3 years ago

Which statement is not correct about force.

BlackZzzverrR [31]

Answer:

c

Explanation:

without force motion won't take place

4 0

2 years ago

All Physicsts over here plz help in these questions!!!!!!!!!

mrs_skeptik [129]

Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.

6 0

3 years ago

Read 2 more answers

The engine in an imaginary sports car can provide constant power to the wheels over a range of speeds from 0 to 70 miles per hou

Mama L [17]

A) Time needed: 6.24 s

B) Time needed: 2.86 s

Explanation:

A)

In this part, we are told that the power if the engine is constant. The power of the engine is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time

This means that the power of the engine is proportional to the work done, and therefore, to the kinetic energy of the car:

$P=\frac{\frac{1}{2}mv^2}{t}=const.$

where m is the mass of the car and v its velocity.

SInce power is constant, we can write:

$\frac{\frac{1}{2}mv_1^2}{t_1^2}=\frac{\frac{1}{2}mv_2^2}{t_2}$

where:

$t_1=1.40 s$ is the time the car needs to accelerates to $v_1=28.0 mph$

$t_2$ is the time the car needs to accelerate to $v_2=57.0 mph$

Therefore, solving for $t_2$ ,

$t_2 = \frac{v^2}{u^2}t_1=\frac{57^2}{28^2}(1.40)=6.24 s$

B)

First of all, we have to calculate the acceleration of the car. We can do it using the following equation:

$a=\frac{v-u}{t}$

where:

u = 0 is the initial velocity

$v=28.0 mph \cdot \frac{1609 m/mi}{3600 s/h}=12.5 m/s$ is the final velocity

t = 1.40 s is the time elapsed

Substituting, we find the acceleration:

$a=\frac{12.5-0}{1.40}=8.9 m/s^2$

In this part, we are told that the force exerted by the engine is constant: according to Newton's second law, acceleration is proportional to the force,

$F=ma$