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3 years ago

Two point charges, 1.8 pC and −1.8 pC, are separated by 7 µm. What is the dipole moment of this pair of charges?

Physics

1 answer:

Salsk061 [2.6K]3 years ago

3 0

Answer: 12,600,000Cm

Explanation:

From the data's;

Charges(q) = 1.8 PC equal to 1.8 x 10^¹²C

Distance = 7 micrometer, is equal to 0.0000070m

From the equation of electric dipole moment, p= q x d, where q= charge, d=distance and p is the dipole moment.

Then we have 1.8x10^¹² x 0.0000070= 12,600,000Cm

NB: The charges are identical.

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The kinetic energy of a ball with the mass of 0.5 kg and a velocity of 10 m/s is what

slamgirl [31]

Answer:

25J

Explanation:

Kinetic energy = 1/2 mv²

Mass = m= 0.5kg

Velocity = v = 10m/s

K. E = 1/2 x 0.5 x 10 x 10

K. E = 0.5 x 5 x 10

K. E = 2.5 x 10

K. E = 25 J

The S. I unit of energy is joules with symbol J

I hope this was helpful, Please mark as brainliest

4 0

3 years ago

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The

nordsb [41]

Answer:

$t=6.96s$

Explanation:

From this exercise, our knowable variables are <u>hight and initial velocity </u>

$v_{oy}=96ft/s$

$y_{o}=112ft$

To find how much time does the <u>ball strike the ground</u>, we need to know that the final position of the ball is y=0ft

$y=y_{o}+v_{oy}t+\frac{1}{2}gt^{2}$

$0=112ft+(96ft/s)t-\frac{1}{2}(32.2ft/s^{2})t^{2}$

Solving for t using quadratic formula

$t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}$

$a=-\frac{1}{2} (32.2)\\b=96\\c=112$

$t=-0.999s$ or $t=6.96s$

<u><em>Since time can't be negative the answer is t=6.96s</em></u>

7 0

3 years ago

Wet sugar that contains one-fifth water by mass is conveyed through an evaporator in which 85% of the of the entering water is e

Natalka [10]

Answer:

i) $v'=\frac{17}{20}$ ii) $\frac{m_v}{m_f-m_v} =\frac{17}{83}$

b) $m_r=752963.55\ kg$

Explanation:

Given:

fraction of water in wet sugar of m kg by mass, $m'_w=\frac{1}{5} \times m$

% of water evapourated from the total water after passing through the evapourator, $m'_v=85\%$

amount of wet sugar fed to the evapourator, $m_f=100\ kg$

Now the mass of water present in the fed amount of sugar:

$m_w=\frac{1}{5} \times m_f$

$m_w=\frac{100}{5}$

$m_w=20\ kg$

Now the amount of water leaving from this total amount of water after passing through the evapourator:

$m_v=m_v' \times m_w$

$m_v=\frac{85}{100} \times 20$

$m_v=17\ kg$

So, the fraction of of water leaving the evapourator:

$v'=\frac{m_v}{m_w}$

$v'=\frac{17}{20}$

ii)

Now the ratio of kg water vaporized/kg wet sugar leaving the evaporator.:

$\frac{m_v}{m_f-m_v} =\frac{17}{100-17}$

$\frac{m_v}{m_f-m_v} =\frac{17}{83}$

amount of sugar fed per day, $m_f=907185\ kg$

<u>Now the mass of water in the given amount of sugar per day:</u>

$m_w=\frac{m_f}{5}$

$m_w=\frac{907185}{5}$

$m_w=181437\ kg$

Mass of water vapourized after passing through the evaporator:

$m_v=\frac{85}{100}\times m_w$

$m_v=\frac{85}{100}\times 181437$

$m_v=154221.45\ kg$

Now the mass of water still remaining in the sugar:

$m_r=m_w-m_v$

$m_r=907185-154221.45$

$m_r=752963.55\ kg$

is the mass of extra water to be evapourated.

8 0

3 years ago

A pendulum is made of a small sphere of mass 0.250 kg attached to a lightweight string 1.20 m in length. As the pendulum swings

forsale [732]

Answer:v=2 m/s

Explanation:

Given

Length of string L=1.2 m

mass of pendulum m=0.25 kg

maximum inclination with vertical \theta =34

vertical Rise of Pendulum from its mean position is given by

$\Delta h=L(1-\cos \theta )$

Conserving Energy at top and bottom point

Potential Energy of sphere is converted into kinetic energy of sphere

$mgL(1-\cos \theta )=\frac{mv^2}{2}$

$v=\sqrt{2gL(1-\cos \theta )}$

$v=\sqrt{2\times 9.8\times 1.2(1-\cos 34)}$

$v=\sqrt{4.021}$

$v=2 m/s$

5 0

3 years ago

A rectangular loop of wire with dimensions 1.80 cm by 9.00 cm and resistance 0.800 Ω is being pulled to the right out of a regio

sladkih [1.3K]

(a) The magnitude of the force that the magnetic field exerts on the loop is 0.042 N.

(b) The direction of the force that the magnetic field exerts on the loop will be out of the plane.

<h3>Magnetic force exerted on the loop</h3>

F = BIL

where;

I is current in the loop
L is length of the loop

emf = BVb

where;

b is breadth of the loop
V is velocity
B is magnetic field

emf = 2.6 x 3 x 0.018 = 0.1404 V

Current in the loop, I = emf/R = 0.1404/0.8 = 0.18 A

<h3>Magnetic force</h3>

F = BIL

where;

L is length of the loop

F = 2.6 x 0.18 x 0.09 = 0.042 N

<h3>Direction of the force</h3>

The direction of the force that the magnetic field exerts on the loop will be out of the plane.

Learn more about magnetic force here: brainly.com/question/13277365

#SPJ1

7 0

2 years ago