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Harman [31]
3 years ago
6

A radio has a 1.3 A current. If it has a resistance of 35 Ω, what is the potential difference?

Physics
2 answers:
DedPeter [7]3 years ago
7 0

Answer:

22

Explanation:

lidiya [134]3 years ago
3 0

Answer:

45.5 v

Explanation:

I = 1.3 A

R = 35 Ω

V = I * R

V = 1.3 * 35 = 45.5 v

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made up of more than one cell.

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Infrared telescopes are usually placed on high-flying airplanes or on satellites in space __________
Yuliya22 [10]

Answer:

In order to improve visibility

Explanation:

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2 years ago
A coil has N turns enclosing an area of A. In a physics laboratory experiment, the coil is rotated during the time interval Δt f
anzhelika [568]

Answer:

\phi_i = BA

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magnetic flux is the count of magnetic field lines passing through a given loop or area

As we know that magnetic flux is given by the formula

\phi = \vec B. \vec A

here we also know that magnetic field B and plane of the coil is perpendicular in initial position

So the area vector is always perpendicular to the plane of the coil

so the angle between magnetic field and area vector is parallel to each other and this angle would be zero

so magnetic flux of the coil initially we have

\phi = BAcos0 = BA

6 0
3 years ago
1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin
aleksandr82 [10.1K]

Answer:

d = 4 d₀o

Explanation:

We can solve this exercise using the relationship between work and the variation of kinetic energy

         W = ΔK

In that case as the car stops v_f = 0

the work is

          W = -fr d

we substitute

          - fr d₀ = 0 - ½ m v₀²

           d₀ = ½ m v₀² / fr

now they indicate that the vehicle is coming at twice the speed

          v = 2 v₀

using the same expressions we find

           d = ½ m (2v₀)² / fr

           d = 4 (½ m v₀² / fr)

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3 0
2 years ago
An electron accelerated from rest through a voltage of 780 v enters a region of constant magnetic field. part a part complete if
maxonik [38]
The electron is accelerated through a potential difference of \Delta V=780 V, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:
\frac{1}{2}mv^2 =  e \Delta V
where
m is the electron mass
v is the final speed of the electron
e is the electron charge
\Delta V is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:
v= \sqrt{ \frac{2 e \Delta V}{m} } = \sqrt{ \frac{2(1.6 \cdot 10^{-19}C)(780 V)}{9.1 \cdot 10^{-31} kg} }=1.66 \cdot 10^7 m/s


Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:
evB=m \frac{v^2}{r}
where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
B= \frac{mv}{er}= \frac{(9.1 \cdot 10^{-31}kg)(1.66 \cdot 10^7 m/s)}{(1.6 \cdot 10^{-19}C)(0.25 m)} =3.8 \cdot 10^{-4} T
3 0
3 years ago
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