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mylen [45]
3 years ago
8

Jeremy has a handful of resistors. All of them, except one, have four bands. The exception has three. Why? A. He should use anot

her because the band may haven't been printed or worn out, and using the resistor could damage the circuit since there's no way to tell its value due to the missing band. B. The resistor has a tolerance of 20%. C. The resistor has a non-decimal value, like 10 ohms, instead of 10.4 ohms. D. The resistor has a low resistance value so it properly only has three bands.
Physics
1 answer:
Art [367]3 years ago
7 0

Answer: option B: The resistor has a tolerance of 20%.

Explanation:

Resistors are color coded to identify their value and function. For a 3- Band resistor, reading from the left, first band denotes first digit, second band depicts second digit and third color depicts the multiplier. The absence of fourth band means the resistor has tolerance 20%.

Hence, the correct answer is option B. The resistor has a tolerance of 20%.

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andreev551 [17]
Physical development , Cognitive Development , psychosocial development I think!
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4 years ago
HELP PLS and HURRY
tangare [24]

Answer:

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Explanation:

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5 0
2 years ago
A particular light source gives off light waves with a measured wavelength of
Umnica [9.8K]

The frequency of the light source is 1.5 x 10¹⁵ Hz.

<h3>Frequency of the light source</h3>

The frequency of the light source is determined using the following equations;

c = fλ

where;

c is speed of light

f is the frequency

λ is the wavelength

f = (3 x 10⁸) / (2 x 10⁻⁷)

f = 1.5 x 10¹⁵ Hz

Thus, the frequency of the light source is 1.5 x 10¹⁵ Hz.

Learn more about frequency of light here: brainly.com/question/10728818

3 0
3 years ago
A large, cylindrical water tank with diameter 3.60 m is on a platform 2.00 m above the ground. The vertical tank is open to the
zysi [14]

To solve this problem it is necessary to apply the concepts related to the geometry of a cylindrical tank and its respective definition.

The volume of a tank is given by

V = \frac{\pi d^2}{4}h

Where

d = Diameter

h = Height

Considering that there are two stages, let's define the initial and final volume as,

V_0 = \frac{\pi d^2}{4}H

V_f = \frac{\pi d^2}{4}h

We know as well by definition that

1gal = 3.785*10^{-3}m^3

Then we have for the statement that

V_f = V_0 -1gal

V_f = V_0 - 3.785*10^{-3}

Replacing the previous data

\frac{\pi d^2}{4}h = \frac{\pi d^2}{4}H- 3.785*10^{-3}

\frac{\pi (3.6)^2}{4}h = \frac{\pi (3.6)^2}{4}(2)- 3.785*10^{-3}

Solving to get h,

h = 1.99963m

Therefore the change is

\Delta h = H-h

\Delta h = 2- 1.99963

\Delta h = 3.7*10^{-4}m=0.37mm

Therefore te change in the height of the water in the tank is 0.37mm

4 0
3 years ago
A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein
RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of 8m^{3}/s. As you  may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

V_{cone}=\frac{1}{3} \pi r^{2}h

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

\frac {r}{h}=\frac{4}{5}

When solving for r, we get:

r=\frac{4}{5}h

so we can substitute this into our volume of a cone formula:

V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h

which simplifies to:

V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h

V_{cone}=\frac{16}{75} \pi h^{3}

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}

Which simplifies to:

\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}

Now we can substitute the provided values into our equation. So we get:

\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}

so:

\frac{dh}{dt}=0.25m/s

3 0
3 years ago
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