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3 years ago

A given 60uc piont charge located at origion

2 answers:

7 0

For my first part it like that as we all know that flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc M I RIGHT PLZ tell me

Morgarella [4.7K]3 years ago

6 0

Answer:

flux density is the charge per unt area here charge is 60uc so divide 60uc by (4*pi*r2) we get D= 7.06*10^-5 c/m2. NOW given is portion with r=26cm theta= 0 to pi/2 and phi = 0 to pi/2 calculate required region area with formula =double integral(r^2sintheta dtheta dphi) we get =.106m^2 now multiply D*Required region we get 7.5uc

Explanation:

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A hypothesis can be accepted as true after ??? repeated trials. *ill give you 10 points

elena55 [62]

Answer:

true

Explanation:

I'm assuming this is a true or false question. if u get the same results after each repeated trial that is the only time a hypothesis can be used to support evidence

7 0

2 years ago

Two point charges are separated by 10 cm, with an attractive force between them of 15 N. Find the force between them when they a

suter [353]

Answer:

(a) the force is 8.876 N

(b) the magnitude of each charge is 4.085 μC

Explanation:

Part (a)

Given;

coulomb's constant, K = 8.99 x 10⁹ N.m²/C²

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

force between the two charges, F = 15 N

when the distance between the charges changes to 13 cm (0.13 m)

force between the two charges, F = ?

Apply Coulomb's law;

$F = \frac{Kq_1q_2}{r^2} \\\\let \ Kq_1q_2 = C\\\\F =\frac{C}{r^2} \\\\C = Fr^2\\\\F_1r_1^2 = F_2r_2^2\\\\F_2 =\frac{F_1r_1^2}{r_2^2} \\\\F_2 = \frac{15*0.1^2}{0.13^2} \\\\F_2 = 8.876 \ N$

Part (b)

the magnitude of each charge, if they have equal magnitude

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

where;

F is the force between the charges

K is Coulomb's constant

Q is the charge

r is the distance between the charges

$F = \frac{KQ^2}{r^2} \\\\Q = \sqrt{\frac{Fr^2}{K} } \\\\Q = \sqrt{\frac{15*(0.1)^2}{8.99*10^9} } = 4.085 *10^{-6} \ C\\\\Q = 4.085 \ \mu C$

4 0

3 years ago

The model below shows a calcium atom. An image has a mix of red and blue balls in its center and 4 concentric black rings around

zubka84 [21]

Answer:

8 electrons in the third energy level

Explanation:

From the description,the third energy level has 8 electron (represented by the small green balls you describe)

4 0

3 years ago

In the two-slit experiment, monochromatic light of wavelength 600 nm passes through a 19) pair of slits separated by 2.20 x 10-5

kumpel [21]

Explanation:

It is given that,

Wavelength of monochromatic light, $\lambda=600\ nm=6\times 10^{-7}\ m$

Slits separation, $d=2.2\times 10^{-5}\ m$

(a) We need to find the angle corresponding to the first bright fringe. For bright fringe the equation is given as :

$d\ sin\theta=n\lambda$ , n = 1

$\theta=sin^{-1}(\dfrac{\lambda}{d})$

$\theta=sin^{-1}(\dfrac{6\times 10^{-7}}{2.2\times 10^{-5}})$

$\theta=1.56^{\circ}$

(b) We need to find the angle corresponding to the second dark fringe, n = 1

So, $d\ sin\theta=(n+\dfrac{1}{2})\lambda$

$sin\theta=\dfrac{3\lambda}{2d}$

$\theta=sin^{-1}(\dfrac{3\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{3\times 6\times 10^{-7}}{2\times 2.2\times 10^{-5}})$

$\theta=2.34^{\circ}$

Hence, this is the required solution.

4 0

3 years ago

An 80-kg man is skating northward and happens to suddenly collide with a 20-kg boy who is ice skating toward the east. Immediate

Fantom [35]

Answer:

6.25 m/s

Explanation:

mass of man (m1) = 80 kg

mass of boy (m2) = 20 kg

mass of man and boy after collision (m12)= 20 + 80 = 100 kg

velocity of man and boy after collision (v) = 2.5 m/s

angle θ = 60 °

How fast was the boy moving just before the collision ?

From the diagram attached, the first image shows the man and the boys motion while the second diagram shows their motion rearranged to form a triangle. With the momentum of the man and the boy forming the sides of the triangle.
M₁₂ = total momentum after collision = m12 x v = 100 x 2.5 = 250
Mboy = momentum of the boy before collision = m2 x Velocity of boy
Mman = momentum of the man before collision = m1 x velocity of man
from the triangle, cos θ = $\frac{Mboy}{M₁₂}$

cos 60 = $\frac{Mboy}{250}$

Mboy = 250 x cos 60 = 125

recall that momentum of the boy (Mboy) also = m2 x Velocity of boy

therefore

125 = 20 x velocity of boy

velocity of boy = 125 / 20 = 6.25 m/s

4 0

3 years ago