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3 years ago

Can someone help me due at 1:00 PM plus 100 points questions B,C,D

Physics

2 answers:

vagabundo [1.1K]3 years ago

6 0

<h3>Hope it helps...</h3>

Alenkasestr [34]3 years ago

5 0

Answer:

B is 1

C is 2 I believe

D I'm pretty sure is 3

Explanation:

you can look the answers up :D!

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Give two differences between Mass and weight

nydimaria [60]

Answer:

my opi

Explanation:

Mass is like the inside of the weight. weight is just home much it weighs

8 0

3 years ago

Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the

IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency $f= 6.37\times10^{14}\ Hz$

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

$\lambda=\dfrac{c}{f}$

where, c = speed of light

f = frequency

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}$

$\lambda=471\ nm$

We need to calculate the distance between the two slits

$m\times \lambda=d\sin\theta$

$d =\dfrac{m\times\lambda}{\sin\theta}$

Where, m = number of fringe

d = distance between the two slits

Here, $\sin\theta =\dfrac{y}{l}$

Put the value into the formula

$d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}$

$d=40.11\times10^{-6}\ m$

$d = 40.11\ \mu m$

Hence, The distance between the two slits is 40.11 μm.

7 0

3 years ago

A person must pull a rope 20 meters with a force of 200 N what is the work input?

tatiyna

Work done = force * distance
work done = 200 * 20
work done = 4000J

4 0

3 years ago

Convierta 164 decimetros a hectometros

-BARSIC- [3]

Answer:

sinco

Explanation:

6 0

3 years ago

Two positive charges ( 8.0 mC and 2.0 mC) are separated by 300 m. A third charge is placed at distance r from the 8.0 mC charge

WINSTONCH [101]

Answer:

$r=200m$

Explanation:

From the question we are told that:

Charges:

$Q_1=8.0mC$

$Q_2=2.0mC$

$Q_3=8.mC$

Distance $d=300m$

Generally the equation for Force is mathematically given by

$F=\frac{kq_1q_2}{r^2}$

Therefore

$F_{32}=F_{31}$

$\frac{q_2}{(300-r)^2}=\frac{q_1}{r^2}$

$\frac{2*10^{-3}}{(300-r)^2}=\frac{8*10^{-3}}{r^2}$

$r=2(300-r)$

$r=200m$

8 0

2 years ago