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3 years ago

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

1 answer:

7 0

I thought you were going to ask for the resistance of the unknown
series resistor. Since you only want the equivalent resistance of the
circuit, you don't even need to know the resistance of the lamp.

I = E / R

Current through the circuit = (voltage of the battery) / (circuit resistance).

0.5 = (12) / R

Multiply each side by 'R' : (0.5) R = 12

Multiply each side by 2 : <em>R = 24 ohms</em>

(Since the resistance of the lamp is 10 ohms, the
unknown series resistor is the other 14 ohms.)

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A 13.0 kg wheel, essentially a thin hoop with radius 1.80 m, is rotating at 469 rev/min. It must be brought to a stop in 16.0 s.

Stella [2.4K]

Answer:

Explanation:

Given

mass of wheel m=13 kg

radius of wheel=1.8 m

N=469 rev/min

$\omega =\frac{2\pi \times 469}{60}=49.11 rad/s$

t=16 s

Angular deceleration in 16 s

$\omega =\omega _0+\alpha \cdot t$

$\alpha =\frac{\omega }{t}=\frac{49.11}{16}=3.069 rad/s^2$

Moment of Inertia $I=mr^2=13\times 1.8^2=42.12 kg-m^2$

Change in kinetic energy =Work done

Change in kinetic Energy $=\frac{I\omega ^2}{2}-\frac{I\omega _0^2}{2}$

$\Delta KE=\frac{42.12\times 49.11^2}{2}=50,792.34 J$

(a)Work done =50.79 kJ

(b)Average Power

$P_{avg}=\frac{E}{t}=\frac{50.792}{16}=3.174 kW$

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3 years ago

What is the relationship between an object’s mass and its gravitational potential energy?.

antiseptic1488 [7]

Answer:

If we’re talking about objects on the Earth, the gravitational potential energy is given by:

Explanation:

PEg=mgh

so the energy is proportional to the mass ( m ), but also to the strength of the gravitational field ( g ), and the height ( h ) to which the mass is lifted.

5 0

2 years ago

The charges and coordinates of two charged particles held fixed in the xy plane are: q1 = +3.3 µc, x1 = 3.5 cm, y1 = 0.50 cm, an

Readme [11.4K]

1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is
$d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$
substituting the coordinates of the two charges, we get
$d= \sqrt{(3.5+2)^2+(0.5-1.5)^2}=5.6~cm=0.056~m$

2) Then, we can calculate the electrostatic force between the two charges $q_1$ and $q_2$ , which is given by
$F=k_e \frac{q_1 q_2}{d^2}$
where $k_e=8.99\cdot10^{9} Nm^2C^{-2}$ is the Coulomb's constant.
Substituting numbers, we get
$F=8.99\cdot10^{9} Nm^2C^{-2} \frac{(3.3\cdot10^{-6}~C) (-4\cdot10^{-6}~C)}{(0.056~m)^2} =-37.8~N$
and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.

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3 years ago

The moon Phobos orbits Mars

podryga [215]

Explanation:

For a circular orbit v= $\sqrt{\frac{G.m}{r} }$ with G = 6.6742 × $10^{-11}$

Given m = 6.42 x 10^23 kg and r=9.38 x 10^6 m

=> v = 2137.3 m/s

I hope this is the correct way to solve

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2 years ago

Explain why the extrapolated temperature is used to determine the maximun temperature of the mixture rather than the highest rec

Oksi-84 [34.3K]

The extrapolated temperature is used to define the maximum temperature of the mixture relatively than the highest recorded temperature in which the conclusion will effect in a higher specific heat value. Heat is bound to escape from whatever apparatus is using, therefore it is needed to account for the loss of the heat that does not go into increasing the temperature of the mixture.

6 0

3 years ago

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