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4 years ago

If the voltage impressed across a circuit is held constant while the resistance is halved, what change occur?

Physics

1 answer:

egoroff_w [7]4 years ago

3 0

Answer:

The current doubles

Explanation:

In a circuit, Ohm's law gives the relationship between voltage, current and resistance:

$V=RI$

where

V is the voltage

R is the resistance

I is the current

In this problem,

V is held constant

R is halved: $R'=\frac{R}{2}$

Therefore, the new current is

$I'=\frac{V}{R'}=\frac{V}{R/2}=2\frac{V}{R}=2I$

So, the current doubles.

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According to Care-stream's operating manual, what is the acceptable range of exposure indicators?

Aleks [24]

Answer:

correct option is C. 1700 - 2300

Explanation:

solution

according to Care stream, exposure indicators provide the technical adequancy incident radiations on the x ray

and we muse care stream formula that is

Exposure Index EI = 1000 × $log_{10} (mR)$ + 2000

here 1 mR = 2000 EI

and 2 mR = 2300 EI

and 3 mR = 1700 EI

so here acceptable range of exposure indicators is 1700 - 2300

correct option is C. 1700 - 2300

5 0

3 years ago

What minimum speed must an electron have to excite the 492-nm-wavelength blue emission line in the hg spectrum?

mezya [45]

The energy required by the excitation of the line is:
ΔE = hν = hc / λ
where:
ΔE = energy difference
h = Planck constant
ν = line frequency
c = speed of light
λ = line wavelength

The energy difference must be supplied by the electron, supposing it transfers all its kinetic energy to excite the line:
$\Delta E = \frac{1}{2} m v^{2}$

Therefore,
$\frac{1}{2} m v^{2} = \frac{hc}{\lambda}$

And solving for v we get:
$v = \sqrt{ \frac{2hc}{m\lambda} }$

Plugging in numbers (after trasforing into the correct SI units of measurement):
$v = \sqrt{ \frac{(2)(6.6 \cdot 10^{-34})(3 \cdot 10^{8}) }{(9.11 \cdot 10^{-31})(4.92 \cdot 10^{-7}) } }$
=9.4 · 10⁵ m/s

Hence, the electron must have a speed of 9.4 · 10<span>⁵ m/s in order to excite the <span>492nm</span> line.</span>

5 0

3 years ago

A truck tire rotates at an initial angular speed of 21.5 rad/s. The driver steadily accelerates, and after 3.50 s the tire's ang

erma4kov [3.2K]

Given:

initial angular speed, $\omega _{i}$ = 21.5 rad/s

final angular speed, $\omega _{f}$ = 28.0 rad/s

time, t = 3.50 s

Solution:

Angular acceleration can be defined as the time rate of change of angular velocity and is given by:

$\alpha = \frac{\omega_{f} - \omega _{i}}{t}$

Now, putting the given values in the above formula:

$\alpha = \frac{28.0 - 21.5}{3.50}$

$\alpha = 1.86 m/s^{2}$

Therefore, angular acceleration is:

$\alpha = 1.86 m/s^{2}$

5 0

3 years ago

How is retrograde motion different from normal diurnal motion

FromTheMoon [43]

Answer:

Diurnal motion, apparent daily motion of the heavens from east to west in which celestial objects seem to rise and set, a phenomenon that results from the Earth's rotation from west to east. The axis of this apparent motion coincides with the Earth's axis of rotation. Retrograde motion is an APPARENT change in the movement of the planet through the sky. It is not REAL in that the planet does not physically start moving backwards in its orbit. It just appears to do so because of the relative positions of the planet and Earth and how they are moving around the Sun.

6 0

4 years ago

Why people won’t want to live near wind turbines.

Tatiana [17]

Answer:

People think they ruin landscapes.

Explanation:

This is a common opinion and will be accepted as a point in any essay.

8 0

3 years ago

Read 2 more answers