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ra1l [238]
3 years ago
9

If u humans were able to have powers what would it be any form make ur own creation of powers

Physics
2 answers:
Iteru [2.4K]3 years ago
8 0
To manipulate time cus u could do a lot w that yk
liraira [26]3 years ago
6 0

Answer:

be any animal cuz why not

You might be interested in
Wavelength is measured in hertz true or false?
g100num [7]
Wavelength is measured in various units of distance (mm, cm, m, etc.). Frequency uses hertz, but that is not what the question is asking. Therefore, the answer is false.
3 0
3 years ago
A charge of -2.65 nC is placed at the origin of an xy-coordinate system, and a charge of 2.00 nC is placed on the y axis at y =
stiks02 [169]

Answer:

A. Fnx = 5.71*10⁻⁵ N  ,  Fny= -3.67*10⁻⁵ N

B. Fn= 6.78 *10⁻⁵ N

C. α= 32.4° counterclockwise with the positive x+ axis

Explanation:

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -2.65 nC =-2.65*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

d_{13} = \sqrt{(3.2)^{2} +(3.8)^{2} }

d_{13} =\sqrt{24.68} * 10⁻²m    = 4.9678* 10⁻²m

(d₁₃)² = 24.68*10⁻⁴m²

d₂₃ = 3.2 cm = 3.2*10⁻²m  

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*2.65*10⁻⁹*5*10⁻⁹) /(24.68*10⁻⁴)

F₁₃ = 4.8 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² =  ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /((3.2)²*10⁻⁴)

F₂₃ = 8.8 *10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -4.8 *10⁻⁵ *cos β= - 4.8 *10⁻⁵(3.2/ (4.9678)= - 3.09*10⁻⁵ N

F₁₃y= -4.8 *10⁻⁵ *sin β= - 4.8 *10⁻⁵(3.8/(4.9678) =  - 3.67*10⁻⁵ N

F₂₃x  = F₂₃ =  +8.8 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x =  - 3.09*10⁻⁵ N+8.8 *10⁻⁵ N= 5.71*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 3.67*10⁻⁵ N+0= - 3.67*10⁻⁵ N

Fn magnitude

F_{n} =\sqrt{(Fn_{x})^{2}+(Fn_{y})^{2}  }

F_{n} = \sqrt{(5.71)^{2}+(3.67)^{2}  } *10⁻⁵ N

Fn= 6.78 *10⁻⁵ N

Fn direction  (α)

\alpha =tan^{-1}( \frac{Fn_{y} }{Fn_{x} } )

\alpha =tan^{-1}( \frac{-3.67 }{5.71} )

α= -32.4°

α= 32.4° counterclockwise with the positive x+ axis

4 0
3 years ago
The electric field of a sinusoidal electromagnetic wave obeys the equation E=(375V/m)cos[(1.99×107rad/m)x+(5.97×1015rad/s)t].a.
andrey2020 [161]

Answer:

A) max = Eo = 375 V/m, B)  B = 125 10⁻⁸ T , C) f = 9.50 101⁴ Hz, D) λ = 3.158 10⁻⁷ m, E)  T = 1.05 10⁻¹⁵ s , F) invisible for humans   G)  v = c = 3 10⁸ m/s

Explanation:

The expression given for the electric field is

       E = 375 cos (1.99 107x + 5.97 105t)

The general formula for the electric field of a transverse traveler is

     E = Eo cos (kx-wt)

Where k is the wave number and w the angular velocity

A) The amplitudes as electric is

     Emax = Eo = 375 V / m

B) the electric and magnetic field are related

     E / B = c

     B = E / c

     B = 375/3 108

     B = 125 10⁻⁸ T

C) angular velocity and frequency is related

    .w = 2π f

     f = w / 2π

     f = 5.97 10¹⁵ / 2π

     f = 9.50 101⁴ Hz

D) the speed of light has the formula

      c = λ f

      λ = c / f

      λ = 3 10⁸ / 9.50 10¹⁴

      λ = 3.158 10⁻⁷ m

E) The period

     T = 1 / f

     T = 1 / 9.5 10¹⁴

     T = 1.05 10⁻¹⁵ s

F) let's reduce the wavelength nm

   λ = 3.158 10⁻⁷ m (10⁹nm / 1m)

   λ = 3.158 10²nm = 315.3 nm

The visible radiation range is between 400nm and 700nm. This radiation is ultraviolet and is invisible humans

G) All electromagnetic radiation has a speed at the speed of light (c)

      v = c = 3 10⁸ m/s

5 0
3 years ago
A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from
Rzqust [24]

The  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

  20 A  (30 - x)↓      x         ↓      20 cm  B 30 cm

                       2N              0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

(30 - x)(2 + 0.9x) = 18

60 + 27x - 2x - 0.9x² = 18

60 + 25x - 0.9x² = 18

0.9x² - 25x - 42 = 0

x = 29.3 cm

Thus, the  distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

Learn more about brainly.com/question/874205 here:

#SPJ1

7 0
2 years ago
Read 2 more answers
Which of the following is a vector?<br> 7 meters<br> 0.007 cm<br> 7x 106m<br> 7 miles Northwest
erma4kov [3.2K]

Answer:

7 miles northeast is the because it has both magnitude and direction .

4 0
3 years ago
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