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anyanavicka [17]
3 years ago
10

what happens to the current across a circuit when the voltage is doubled while the resistance is held back ​

Physics
2 answers:
babunello [35]3 years ago
8 0
The current is directly proportional to the voltage and inversely proportional to the resistance. So doubling or tripling the resistance will cause the current to be one-half or one-third the original value
Lynna [10]3 years ago
7 0

Since the current is inversely propotional to its resistance, when the voltage is doubled the current will be one-half

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What is kinematics?<br>explain!!~​<br><br>thankyou ~
Hatshy [7]

Answer:

<u>Kinematics:-</u>

  • Branch of mechanics concerned
  • Features of motion in an object
  • Causes and effects of motions

Hope it helps!

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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

Find:

Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
4 years ago
Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


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