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Natali5045456 [20]
3 years ago
13

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

he electric force? The constant k = 9 x 109 N-m2/C2
Physics
1 answer:
tiny-mole [99]3 years ago
6 0

Explanation:

Charges,q_1=8\ \mu C=8\times 10^{-6}\ C

q_2=-5\ \mu C=-5\times 10^{-6}\ C

The distance between charges, r = 10 cm = 0.1 m

We need to find the magnitude and direction of the electric force. It is given by :

F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times 8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\\\F=36\ N

So, the required force between charges is 36 N and it is towards positive charge i.e. +8 μC.

You might be interested in
H E L P ! ! !
madam [21]
The Correct answer to number 1 is A or D

The correct answer to number 2 is C because transmission meaning is to travel.

PS. I think number 1 is D
3 0
3 years ago
(a) Consider a fusion generator built to create 3.00 GW of power. Determine the rate of fuel burning in grams per hour if the DT
Stels [109]

The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 ×10^1^3 J/g per hour

<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
  • D + T → He + n
  • The D-T fusion reaction results in a Helium (He) and  neutron (n)

E = 17.59 MeV

Mass = 2.014u + 3.016u

= 5.030u

Energy per Kg = (17.59×10^6×1.6×10^-^1^9) ÷ ( 5.030×1.66×10^-^2^7)

= 3.37×10^1^4 J/Kg

= 3.0× 10^9 J/g

Rate of fuel burning in grams per hour = 3.0× 10^9 ×  3600

= 3.6×3.0×10^1^2

= 1.08 ×10^1^3 J/g per hour

To learn more about fusion reactor and energy production, refer

brainly.com/question/13399644

#SPJ4

4 0
2 years ago
A stone is thrown vertically upward with velocity 9.8ms 1)calculate the time taken to reach maximum height.2)what is the maximum
ira [324]

Answer:

1.t=-1.96sec

2.H=4.8m

3.T=1.96sec

4.R=19.2m

Explanation:

u=9.8,t=?,sin theta=1

using formula t=2usintheta/g

t=2x9.8x1=19.6/10

t=1.96seconds

using formula H=u(squared)sin(squared)theta/2g

H=9.8(squared)x1(squared)/2x10

H=96x1/20

H=96/20

H=4.8m

using formula T=2usintheta/g

T=2x9.8x1/10

T=19.6/10

T=1.96sec

using the formula R=u(squared)sin2theta/g

R=9.8(squared)x2/10

R=96x2/10

R=192.08/10

R=19.2

5 0
3 years ago
Read 2 more answers
Plz help asap.
stich3 [128]

Answer:

In a primary cell, two electrodes (one of copper and other of zinc) of metal atoms are used. These electrodes are dipped in an electrolyte solution that causes the metals to produce their respective positive and negative ions.

In this way, the flow of charges takes place and supply the electricity to the source.

Unlike a primary cell, a dry cell contains paste of an electrolyte instead of the solution. The contents of electrolyte paste react with each other through a chemical process and convert the chemical energy into electrical energy.

5 0
3 years ago
(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is u
stiv31 [10]

Answer:

(a) 2.5 cm

(b) Yes

Solution:

As per the question:

Mass of Uranium-235 ion, m = 3.95\times 10^{- 25}\ kg

Mass of Uranium- 238, m' = 3.90\times 10^{- 25}\ kg

Velocity, v = 3.00\times 10^{5}\ m/s

Magnetic field, B = 0.250 T

q = 3e

Now,

To calculate the path separation while traversing a semi-circle:

\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})

The radius of the ion in a magnetic field is given by:

R = \frac{mv}{qB}

\Delta x = 2(R_{U_{35}} - 2R_{U_{38}})

\Delta x = 2(\frac{mv}{qB} - \frac{m'v}{qB})

\Delta x = 2(\frac{m - m'}{qB}v)

Now,

By putting suitable values in the above eqn:

\Delta x = 2(\frac{3.95\times 10^{- 25} - 3.90\times 10^{- 25}}{3\times 1.6\times 10^{- 19}\times 0.250}\times 3.00\times 10^{5}) = 2.5\ cm

\Delta x = 1.25\ cm

(b) Since the order of the distance is in cm, thus clearly this distance is sufficiently large enough in practical for the separation of the two uranium isotopes.

3 0
3 years ago
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