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ycow [4]
3 years ago
14

The force that moving charged particles exert on one another is called

Physics
1 answer:
Irina-Kira [14]3 years ago
6 0
It should be electromagnetic force
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Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x
GalinKa [24]

Answer:

a) (-367231.63i ,  367231.63i, 0) N/C

b) (0 , 0  , 367231.63i ) N/C

Explanation:

a)

Case x < -2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i

Case x > 2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

Case -2.15 < x <+2.15

E =E_{+} - E_{+} \\\\E = 0

b)

Case x < -2.15

E =E_{+} - E_{+} \\\\E = 0

Case x > 2.15

E =E_{+} - E_{+} \\\\E = 0

Case -2.15 < x <+2.15

E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

7 0
3 years ago
An ecology center wants to set up an experimental garden using 400 meters of fencing to enclose a rectangular area of 9900 squar
blsea [12.9K]

Explanation:

Given that,

Fencing to be done, P = 400 m

Area of the rectangular garden, A=9900\ m^2

Let l is the length and b is the breadth of the garden. The area of the rectangle is given by :

A=l\times b

lb=9900............(1)

The perimeter of the garden is given by :

P=2(l+b)

400=2(l+b)

200=(l+b)...............(2)

On solving equation (1) and (2), we get the value of l and b are as follows :

l = 90 m and b = 110 m

or

b = 90 m and l = 110 m

So, the dimension of the garden is 90 m by 110 m. Hence, this is the required solution.

6 0
3 years ago
What is the chemical symbol for C6H12O6
Flura [38]
The Answer is D-Glucose.
5 0
4 years ago
Read 2 more answers
Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i
romanna [79]

Answer:

v_f=0.825m/s

Explanation:

We must use conservation of linear momentum before and after the collision, p_i=p_f

Before the collision we have:

p_i=p_1+p_2=m_1v_1+m_2v_2

where these are the masses are initial velocities of both players.

After the collision we have:

p_f=(m_1+m_2)v_f

since they clong together, acting as one body.

This means we have:

m_1v_1+m_2v_2=(m_1+m_2)v_f

Or:

v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Which for our values is:

v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s

7 0
4 years ago
If you weigh 665 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun a
77julia77 [94]

Answer:

The weight on the neutron star is 99.69 × 10^{12} N.

Explanation:

Here,

weight on earth is 665 N. We can calculate the mass as :

                                                 W_{earth}= mg

                                           or, 665= m × 9.810

                                              ∴ m = \frac{665}{9.810} = 67.68 kg

Now weight of this mass on the surface of neutron star:

                            W_{neutron star} = \frac{GMm}{R^{2} }

                                                = \frac{(6.674*10^{-11})(1.99*10^{30})(67.68) }{9500^{2} }

                                                = 99.69 × 10^{12} N

The required weight is 99.69 × 10^{12} N.

7 0
3 years ago
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