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3 years ago

A 1.5V battery is connected to a small light bulb with a resistance of 3.5 what is the current in the bulb?

1 answer:

5 0

Technically, we don't have the information needed to calculate the current,
because you haven't mentioned the units of the 3.5 .

Since the 3.5 is a resistance, we strongly suspect ... and we'll therefore
assume ... that the 3.5 has the units of ohms. Then . . .

Current = (voltage) / (resistance) = (1.5/3.5) = 3/7 of an Ampere.

(429 mA, rounded)

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What does a measured number tell you?

oksano4ka [1.4K]

Answer:

The precision at which the number was measured

Explanation:

The number of digits that you use to record a measurement indicates the precision of the measurement.

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4 years ago

When an object is located 32 cm to the left of the lens, the image is formed 17 cm to the right of the lens. What is the focal l

Harman [31]

Answer: 11.1cm

Explanation:

Object distance (u) = 32cm

Image distance(v) = 17cm

Focal length(f) =?

Lens formular:

1/f = 1/u + 1/v

1/f = 1/32 + 1/17

Taking the L. C. M of 17 and 32

1/f = (17 + 32) / 544

1/f = 49/544

Taking the reciprocal

f = 544/49

f = 11.1 cm

Therefore, Focal length of the lens is 11.1cm

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4 years ago

What is the pressure drop due to the bernoulli effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Paul [167]

Answer:

$\Delta P=1581357.92\ Pa$

Explanation:

Given:

diameter of hose pipe, $D=0.09\ m$

diameter of nozzle, $d=0.03\ m$

volume flow rate, $\dot{V}=40\ L.s^{-1}=0.04\ m^3.s^{-1}$

Now, flow velocity in hose:

$v_h=\frac{\dot V}{\pi.D^2\div 4}$

$v_h=\frac{0.04\times 4}{\pi\times 0.09^2}$

$v_h=6.2876\ m.s^{-1}$

Now, flow velocity in nozzle:

$v_n=\frac{\dot V}{\pi.d^2\div 4}$

$v_n=\frac{0.04\times 4}{\pi\times 0.03^2}$

$v_n=56.5884\ m.s^{-1}$

We know the Bernoulli's equation:

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}+Z_1=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}+Z_2$

when the two points are at same height then the eq. becomes

$\frac{P_1}{\rho.g}+\frac{v_1^2}{2g}=\frac{P_2}{\rho.g}+\frac{v_2^2}{2g}$

$\Delta P=\frac{\rho(v_n^2-v_h^2)}{2}$

$\Delta P=\frac{1000(56.5884^2-6.2876^2)}{2}$

$\Delta P=1581357.92\ Pa$

8 0

3 years ago

What is the current when a typical static charge of 0.234~\mu\text{c}0.234 μc moves from your finger to a metal doorknob in 0.59

Black_prince [1.1K]

The current is defined as the quantity of charge Q that passes through a certain location in a time $\Delta t$ :
$I= \frac{Q}{\Delta t}$
Using the data of the problem, we find:
$I= \frac{Q}{\Delta t} = \frac{0.234 \mu C}{0.595 \mu s}=\frac{0.234\cdot 10^{-6} C}{0.595 \cdot 10^{-6} s}=0.39 A$

4 0

3 years ago

When an electric drill is used to bore a hole into a wooden beam, the user notices that the drill bit gets very hot.

Bess [88]

Wood is a bad conductor of heat. The drilling process involve heat generation between the drill and the wooden material that can not be dissipated away easily. With deep drilling the drill may get so hot that the chips come out in a state of smoking coal.

7 0

3 years ago