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2 years ago

PLEASE HELP WITH GIVE BRAINLIEST!!! ❤️

1 answer:

6 0

The force that the book exerts on the table is a normal force, not a weight force. (The book's weight doesn't act on the table, it acts on the book.) It's equal in magnitude to the weight of the book, again, because of the first law.

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An emf is induced in response to a change in magnetic field inside a loop of wire. Which of the following changes would increase

goldenfox [79]

Answer:

changing the magnetic field more rapidly

Explanation:

According to Faraday's law, whenever there is a change in the magnetic lines of force, it leads the production of induced emf. The magnitude of induced emf is proportional to to the rate of change of flux.

Hence if the magnetic field inside a loop of wire is changed rapidly, the magnitude of induced emf increases in accordance with Faraday's law of electromagnetic induction stated above when the magnetic field is changed more rapidly, hence the answer.

8 0

3 years ago

What kind of frequency do radio waves have?<br><br>A. High frequency <br><br>B. Low frequency

inn [45]

B low frequency it is the lowest frequency

7 0

3 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

3 years ago

A mass is bobbing up and down on a spring. If you increase the amplitude of the motion, how does this affect the time for one os

NikAS [45]

Answer:

It will not change

Explanation:

The period of oscillation of a mass-spring system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

where

T is the period

m is the mass hanging on the spring

k is the spring constant

As we see from the formula, the period of oscillation does not depend on the amplitude of the motion: therefore, if we change the amplitude, the time for one oscillation will not change.

6 0

3 years ago

Which of the following cannot be used to describe a longitudinal wave?

alex41 [277]

Answer:

crest

hope It helped you!

8 0

3 years ago