Ask question

Ask question

All categories

3 years ago

14

Which of the following are not units of frequency: Hertz. m/s. waves/s

1 answer:

ra1l [238]3 years ago

8 0

The answer is waves.

You might be interested in

What has to happen for a force to do work on an object?

dexar [7]

It has to move through a distance over time.

4 0

2 years ago

A thin, metallic spherical shell of radius 0.227 m has a total charge of 6.03 × 10 − 6 C placed on it.

KATRIN_1 [288]

Answer:

Explanation:

Given

radius $r=0.227 m$

Charge on surface $Q=6.03\times 10^{-6} C$

Point Charge inside sphere $q=1.15\times 10^{-6} C$

Electric Field at $r=0.735 m$

Treating Surface charge as Point charge and applying Gauss law

$E_{total}A=\frac{q_{enclosed}}{\epsilon _0}$

where A=surface area up to distance r

$E_{total}=\frac{Q+q}{4\pi r^2}$

$E_{total}=\frac{6.03\times 10^{-6}+1.15\times 10^{-6}}{4\pi (0.735)^2\times 8.85\times 10^{-12}}$

$E_{total}=1.194\times 10^{5} N/C$

3 0

3 years ago

Plz help urgent <br> i will give brainliest

In-s [12.5K]

Answer:

The answer is D. Balanced forces

Hope it helps.......... pls mark as brainliest

5 0

2 years ago

A student has designed a circuit with one battery, two light bulbs, and two switches, pictured to the left. She would like for e

Gnoma [55]

Answer:

The bulb is not powered because the negative end of the battery is not connected to the electric current,making it dysfunctional. The electrons have to flow through the positive and negative ends of the battery in order for it to work.

Hope this helps.

5 0

2 years ago

Read 2 more answers

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago