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jek_recluse [69]
3 years ago
7

A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of th

e force exerted on Q by q
Physics
1 answer:
Anon25 [30]3 years ago
8 0

Answer:

0.3 N

Explanation:

Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.

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If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y directio
alexandr1967 [171]

Answer:

Explanation:

The direction of propagation of electromagnetic wave

is given by the direction of vector E x B where E is electrical field , B is magnetic field .

Given Electric field  = E i because it is along x axis

Magnetic field = Bj because it is along y axis

E x B = Ei x Bj

= EB k .

so direction of E  x B is along k direction or z  - axis so wave is propagating along z - axis .

7 0
3 years ago
Which of the following is an example of technology influencing science?
FinnZ [79.3K]

Answer:A and C

Explanation:

8 0
3 years ago
How do I solve such problem???
pashok25 [27]

As far as I'm concerned, this is a bogus question, or at least a severely corrupted one.

The three numbers given can NOT all be true on Earth.

-- It rolled off the table at 7.6 m/s .  By golly, there you are!  Its initial horizontal velocity is 7.6 m/s, and it has no vertical velocity until it leaves the table.

-- There are no horizontal forces that we're aware of acting on the object.  So it maintains the same horizontal velocity for the rest of the story.  It's 10.5m away from the table in (10.5 m) / (7.8 m/s) = 1.35 second .

-- Vertically, it's just an object dropped from 17.6m off the floor.  Shockingly, the distance it falls in time 'T' is (1/2 g) T².  In 1.35 second, that's 8.88 meters ! . . . only about halfway to the floor !

-- In order to fall 17.6 m to the floor, it would need 1.89 seconds.  In <u>that</u> length of time, however, it would travel (7.8 m/s) x (1.89 s) = 14.78 m away from the base of the table.

So you see, either . . .

-- the table is NOT 17.6m tall, or

-- the object does NOT roll off of the table at 7.8 m/s, or

-- it does NOT land 10.5 m away from the base of the table.

OR . . .

-- the table is not on Earth, and gravity is not 9.8 m/s² !

We often see questions posted on Brainly with not enough given information, OR with some information given that's not needed because it's not involved the answer.  

THIS one is different, and it's unusual.  In this one, we have<em> too much</em> given information, we can't ignore any of it because it's all related, but it's inconsistent and it CAN't all be true.

(Unless the whole story takes place on a mystery planet that is not Earth.  Which I'm not going to take the time and effort right now to figure out what the acceleration of gravity has to be in order to make all of the given information compatible.)

7 0
3 years ago
a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for
Mrrafil [7]
using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
3 0
3 years ago
Two ropes have equal length and are stretched the same way. The speed of a pulse on rope 1 is 1.4 times the speed on rope 2. Par
kondor19780726 [428]

Answer:

m1/m2 = 0.51

Explanation:

First to all, let's gather the data. We know that both rods, have the same length. Now, the expression to use here is the following:

V = √F/u

This is the equation that describes the relation between speed of a pulse and a force exerted on it.

the value of "u" is:

u = m/L

Where m is the mass of the rod, and L the length.

Now, for the rod 1:

V1 = √F/u1 (1)

rod 2:

V2 = √F/u2 (2)

Now, let's express V1 in function of V2, because we know that V1 is 1.4 times the speed of rod 2, so, V1 = 1.4V2. Replacing in the equation (1) we have:

1.4V2 = √F/u1 (3)

Replacing (2) in (3):

1.4(√F/u2) = √F/u1 (4)

Now, let's solve the equation 4:

[1.4(√F/u2)]² = F/u1

1.96(F/u2) =F/u1

1.96F = F*u2/u1

1.96 = u2/u1 (5)

Now, replacing the expression of u into (5) we have the following:

1.96 = m2/L / m1/L

1.96 = m2/m1 (6)

But we need m1/m2 so:

1.96m1 = m2

m1/m2 = 1/1.96

m1/m2 = 0.51

5 0
4 years ago
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