Ask question

Ask question

All categories

4 years ago

10

A flashlight is powered by 4, 1.5-volt batteries, which are connected in series. The bulb draws 720mA or 0.72 amps when turned o

n. Calculate the resistance of the bulb.

1 answer:

Varvara68 [4.7K]4 years ago

3 0

R=0.12
Use the equation V=I/R
V= voltage (volts)
I= current (amps)
R= resistance (ohms)

You might be interested in

A child moving at constant velocity carries a 2 N ice-cream cone 1 m across a level surface. What is the net work done on the ic

VMariaS [17]

Answer:

2 Joule

Explanation:

Work=force *dISPLACMENT

2N*1M

2 JOUL

4 0

3 years ago

When the plutonium bomb was tested in New Mexico in 1945, approximately 1 gram of matter was converted into energy. Suppose anot

gayaneshka [121]

Answer:

The value is $E = 1.35 *10^{14} \ J$

Explanation:

From the question we are told that

The mass of matter converted to energy on first test is $m = 1 \ g = 0.001 \ kg$

The mass of matter converted to energy on second test $m_1 = 1.5 \ g = 1.5 *10^{-3} \ kg$

Generally the amount of energy that was released by the explosion is mathematically represented as

$E = m * c^2$

=> $E = 1.5 *10^{-3} * [ 3.0 *10^{8}]^2$

=> $E = 1.35 *10^{14} \ J$

7 0

3 years ago

Find the mass of the solid cylinder Dequals{(r,theta,z): 0less than or equalsrless than or equals2, 0less than or equalszless

anygoal [31]

The mass of the cylinder <em>D</em> is obtained by integrating the density function over <em>D</em>:

$\displaystyle\iiint_D\rho(r,\theta\,z)\,\mathrm dV$

With $\rho(r,\theta,z)=1+\frac z2$ , and

$D=\left\{(r,\theta,z)\mid 0\le r\le2,0\le z\le10\right\}$

the mass would be

$\displaystyle\int_0^{2\pi}\int_0^2\int_0^{10}1+\frac z2\,\mathrm dz\,\mathrm dr\,\mathrm d\theta=4\pi\int_0^{10}1+\frac z2\,\mathrm dz$

$4\pi\left(z+\dfrac{z^2}4\right)\bigg|_0^{10}$

$=4\pi\left(10+\dfrac{100}4\right)=\boxed{140\pi}$

6 0

3 years ago

How do motion and Newton's laws apply to your everyday life?

Marina86 [1]

When you jumó ur legs put force on ground

6 0

3 years ago

A stereo speaker is rated at P1000 = 52 W of output at 1000 Hz. At 20 Hz, the sound intensity level LaTeX: \betaβ decreases by 1

baherus [9]

Answer:

The value of the power is $P_c = 38.55 \ W$

Explanation:

From the question we are told that

The power rating $P_{1000} =P_b= 52 \ W$

The frequency is $f = 1000 \ Hz$

The frequency at which the sound intensity decreases $f_k = 20 \ Hz$

The decrease in intensity is by $\beta = 1.3 dB$

Generally the initial intensity of the speaker is mathematically represented as

$\beta_1 = 10 log_{10} [\frac{P_b}{P_a} ]$

Generally the intensity of the speaker after it has been decreased is

$\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]$

So

$\beta_1-\beta_2 = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]$

=> $\beta = 10 log_{10} [\frac{P_c}{P_a} ]- 10 log_{10} [\frac{P_b}{P_a} ]= 1.3$

=> $\beta =10log_{10} [\frac{\frac{P_b}{P_a}}{\frac{P_c}{P_a}} ] = 1.3$

=> $\beta =10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $10log_{10} [\frac{P_b}{P_c} ] = 1.3$

=> $log_{10} [\frac{P_b}{P_c} ] = 0.13$

taking atilog of both sides

$[\frac{P_b}{P_c} ] = 10^{0.13}$

=> $[\frac{52}{P_c} ] = 10^{0.13}$

=> $P_c = \frac{52}{1.34896}$

=> $P_c = 38.55 \ W$

3 0

3 years ago