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Pavel [41]
3 years ago
6

A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the c

ircuit?
Physics
2 answers:
UkoKoshka [18]3 years ago
8 0

Answer:0.00125 watts

Explanation:

resistance=50 ohms

Current=5 milliamps

Current=5/1000 milliamps

Current =0.005 amps

power=(current)^2 x (resistance)

Power=(0.005)^2 x 50

Power=0.005 x 0.005 x 50

Power=0.00125 watts

inessss [21]3 years ago
8 0

Answer:

0.00125 Watts

Explanation:

P = I²R

P = (5/1000)² × 50

P = 1.25 × 10^-3 Watts

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a disk with a radius of 0.1 m is spinning about its center with a constant angular speed of 10 rad/sec. What are the speed and m
padilas [110]

The speed of the bug at the rim of the circular disk is \boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}} and the acceleration of the bug is \boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}.

Further Explanation:

Given:

The angular velocity of the disk is 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}.

The radius of the circular disk is 0.1\,{\text{m}}.

Concept:

The linear speed of the bug present at the rim of the circular disk is given by.

v = r \times \omega  

Here, v is the linear speed,r is the radius of the disk and \omega is the angular speed of the disk.

Substitute 0.1\,{\text{m}} for r and 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for \omega in above expression.

\begin{aligned}v &= 0.1\,{\text{m}} \times 10\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\\end{aligned}  

The magnitude of the centripetal acceleration of the bug cling to the rim of the disk is given as.

{a_c} = \dfrac{{{v^2}}}{r}  

Substitute 1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for v and 0.1\,{\text{m}} for r in above equation.

\begin{aligned}a_c&=\dfrac{(1\text{m/s})^2}{0.1\text{m}}\\&=\frac{1}{0.1}\text{m/s}^2\\&=10\,\text{m/s}^2\end{aligned}  

Thus, the speed of the bug at the rim of the circular disk is \boxed{1\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}} and the acceleration of the bug is \boxed{10\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}.

Learn More:

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  2. Max and Maya are riding on a merry-go-round that rotates at a constant speed. If the merry-go-round makes 3.5 revolutions in 9.2 seconds brainly.com/question/8444623
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Answer Details:

Grade: College

Chapter: Uniform Circular motion

Subject: Physics

Keywords:  Circular disk, circular motion, angular speed, linear speed, bug clinging, rim of the disk, acceleration, magnitude, constant, rad/s.

7 0
3 years ago
Read 2 more answers
The coefficient of kinetic friction between Gary's shoes and the wet kitchen floor is 0.09. If Gary has a mass of 65 kg, what is
mariarad [96]

Answer:

57 N

Explanation:

Draw a free body diagram.  There are three forces:

Weight force mg pulling down.

Normal force N pushing up.

Friction force F pushing horizontally.

Sum of the forces in the y direction:

∑F = ma

N − mg = 0

N = mg

Friction force is the product of normal force and coefficient of friction:

F = Nμ

F = mgμ

F = (65 kg) (9.8 m/s²) (0.09)

F = 57.3 N

Rounded, the friction force is 57 N.

8 0
3 years ago
A student is given a sample of matter and asked to determine whether it is an element.
Pepsi [2]
I think the correct answer is C
5 0
3 years ago
Read 2 more answers
A square block of steel with volume 10 cm3 and mass of 75 g is cut precisely in half. The density of the two smaller pieces is n
dem82 [27]
A. Density only depends on the substance. It doesn't matter whether you have a little chip of it or a supertanker full of it ... the density doesn't change.
4 0
3 years ago
Read 2 more answers
If the spring constant is doubled, what value does the period have for a mass on a spring?
quester [9]

Answer:

The period would decrease by `sqrt(2)`.

Explanation:

The angular frequency omega of an oscillating mass m due to a spring with a constant k is given by

\omega = \sqrt{\frac{k}{m}}

(this is obtained by solving the differential equation m\ddot{{ x}}-kx=0)

If k doubles, i.e., k'=2k, then

\omega'=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{m}}=\sqrt{2}\sqrt{\frac{k}{m}}=\omega\sqrt{2}

Since the angular frequency is \omega = \frac{2\pi}{T}, we can say that

\omega\sqrt{2}=\frac{2\pi}{T}\sqrt{2}=\frac{2\pi}{\frac{T}{\sqrt{2}}}=\frac{2\pi}{T'}

and so it becomes clear that the period T will decrease by sqrt(2) as stated in choice (D).

5 0
3 years ago
Read 2 more answers
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