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3 years ago

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A uniform rectangular segment of copper wire with length and cross-sectional area has resistance: R. Then this resistor is melte

d down, and 1/5 the material is used to make a new uniform rectangular resistor with resistance R. What is the length of thisnew resistor?

1 answer:

erica [24]3 years ago

4 0

Answer:

Explanation:

Let L and A be the length and area of cross section of rectangular segment

If resistor is melted down to and its $\frac{1}{5}$ material is used to make a new uniform rectangular resistor with same resistance R

volume of original rectangular(V) $=A\times L$

Volume of new rectangular segment with length $L_0(V_0)=A\times L_0$

$5V_0=V$

$5\times A_0\times L_0=A\times L$ ----1

Resistance is same

$R=\frac{\rho L}{A}=\frac{\rho L_0}{A_0}$

$\frac{L}{L_0}=\frac{A}{A_0}$

substitute $\frac{L}{L_0}$ in 1

$5=\frac{L^2}{L_0^2}$

$\frac{L}{L_0}=\sqrt{5}$

$L=\sqrt{5}L_0$

$L_0=\frac{L}{\sqrt{5}}$

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A rectangular storage container with open top is to have a volume of 50 ft3 . The length of the base of the container is to be t

spayn [35]

Answer:

Length = 2.862 ft

Width = 1.431 ft

Height = 3.05 ft

Explanation:

Let L be lenght, W be width, and H be height of the storage container. Since length is to be twice the width, it means L = 2W. The volumn of the container would also be:

$V = LWH = L(2L)H = 2L^2H$

SInce the volumn constraint is 50, that means:

$2L^2H = 50$

$H = \frac{25}{L^2}$

The cost for the base would be its area times unit price

$C_b = A_bU_b = LWU_b = 2L^2U_b = 2*8L^2 = 16L^2$

Likewise, the cost for the sides would be

$C_s = A_sU_s = 2(L+W)HU_s = 6LH5 = 30LH$

We can substitute H from the equation above:

$C_s = 30L\frac{25}{L^2} = \frac{750}{L}$

Therefore the total material cost

$C = C_b + C_s = 16L^2 + \frac{750}{L}$

To find the minimum of this function, we can take first derivative then set to 0:

$C^{'} = (16L^2)^{'} + (\frac{750}{L})^{'} = 0$

$32L - \frac{750}{L^2} = 0$

$32L = \frac{750}{L^2}$

$L^3 = 750/32 = 23.4375$

$L = 23.4375^{1/3} \approx 2.862ft$

So W = L/2 = 1.43ft and

$H = \frac{25}{L^2} = 3.05 ft$

If we take the 2nd derivative and substitute L = 2.862 we would have

$C{''} = 32 + 2\frac{750}{L^3} = 32 + \frac{1500}{23.43} = 96 > 0$

Hence L = 2.862 would yield the minimum material cost

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3 years ago

A standard light bulb emits light rays in _________ directions. A. parallel B. opposite C. perpendicular D. nearly all

Lena [83]

Well the answer is D nearly all because light bulbs are roundish

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3 years ago

Which applications, either for diagnostic purposes or for therapeutic purposes, involve radioactive materials placed in the body

CaHeK987 [17]

When a radioactive material is required to be placed in the body, the applications are brachytherapy and radioisotope imaging.

Radioactive materials are elements which has the ability to disintegrate by emitting radioactive substance or radiation. A good example of this is Cobalt-60, Titanium-99 etc.

Brachytherapy is a therapeutic process in which radioactive material is inserted into the body in close proximity to the region affected. The radioactive material emits radiations which are required to control the unwanted biological material in the body. A good application of this is the treatment of cancer using Cobalt-60.

Radioisotope imaging is a diagnostic process which is an imaging technique that may require placing a radioactive material in the body so as to trace or locate the affected part of the body. In this case, the material is used as a tracing element.

The applications that require the placing of radioactive materials in the body are brachytherapy and radioisotope imaging.

For more explanation, visit: brainly.com/question/9790340

5 0

3 years ago

Read 2 more answers

Priscilla is driving her car on a busy street and Harvey passes her on his motorcycle. What will happen to the sound from his mo

Marrrta [24]

Answer:

what will happen to the sound is the wavelength increases thus frequency decreases so the sound decreases as he move away from her car..

6 0

3 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago