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Dmitry [639]
3 years ago
9

Two obiect accumulated a charge of

Physics
1 answer:
tamaranim1 [39]3 years ago
3 0

Answer:

A. 181.24 N

Explanation:

The magnitude of hte electrostatic force between two charged objects is given by the equation

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

In this problem, we have:

q_1=4.5\mu C=4.5\cdot 10^{-6}C is the magnitude of the 1st charge

q_2=2.8\mu C=2.8\cdot 10^{-6}C is the magnitude of the 2nd charge

r = 2.5 cm = 0.025 m is the separation between the charges

Therefore, the magnitude of the electric force is:

F=\frac{(9\cdot 10^9)(4.5\cdot 10^{-6})(2.8\cdot 10^{-6})}{(0.025)^2}=181.44 N

So, the closest answer is

A) 181.24 N

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Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface
nekit [7.7K]

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

sin\theta_c=\frac{n_2}{n_1}

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30

hope this helps!!

6 0
3 years ago
Nếu tăng khoảng cách giữa hai điểm lên 4 lần thì lực tương tác tĩnh điện giữa chúng sẽ
ohaa [14]

Answer:

điện giữa chúng s

Explanation:

8 0
2 years ago
The force of gravity on a 2-kg rock is twice as great as that on a 1-kg rock. why then doesn't the heavier rock fall faster?
katrin2010 [14]
It takes twice the force to produce the same acceleration in the 2kg rock. 
5 0
3 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
3 years ago
An object weighs 79.1 N in air. When it is suspended from a force scale and completely immersed in water the scale reads 21.8 N.
Free_Kalibri [48]

Answer:

(a). The density of the object is 1382 kg/m³.

(b). The density of the oil is 536.4 kg/m³.

Explanation:

Given that,

Weight in air = 79.1 N

Weight in water = 21.8 N

Weight in oil = 48.4 N

We need to calculate the volume of object

Using formula of buoyant force

F_{b}=W_{air}=W_{water}

F_{b}=79.1-21.8

F_{b}=57.3\ N

F_{b}=\rho g h

Put the value into the formula

57.3=1000\times V\times 9.8

V=\dfrac{57.3}{1000\times9.8}

V=5.84\times10^{-3}\ m^3

We need to calculate the density

Using formula of buoyant force

F_{b}=\rho Vg

79.1=\rho\times5.84\times10^{-3}\times9.8

\rho=\dfrac{79.1}{5.84\times10^{-3}\times9.8}

\rho=1382\ kg/m^3

The density of the object is 1382 kg/m³.

(b). We need to calculate the volume of object

Using formula of buoyant force

F_{b}=W_{air}=W_{oil}

F_{b}=79.1-48.4

F_{b}=30.7\ N

We need to calculate the density

Using formula of buoyant force

F_{b}=\rho_{oil} Vg

30.7=\rho_{oil}\times5.84\times10^{-3}\times9.8

\rho_{oil}=\dfrac{30.7}{5.84\times10^{-3}\times9.8}

\rho=536.4\ kg/m^3

The density of the oil is 536.4 kg/m³.

Hence, This is the required solution.

4 0
3 years ago
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