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4 years ago

What is the usual time ice takes to evaporate above a Bunsen burner?

Physics

1 answer:

Step2247 [10]4 years ago

5 0

Depends on the volume but a normal ice cube takes less than 20 seconds

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Would a decrease in force cause a decrease in pressure

natta225 [31]

IdkAnswer:

Explanation:

4 0

4 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

Water floats on a liquid called "carbon tetrachloride." The two liquids do not mix. A light ray passing from water into carbon t

Oduvanchick [21]

Answer:

1.26

Explanation:

$index \: of \: refraction \: = \: \ \frac{ \sin(angle \: of \: incidence) }{ \sin(angle \: of \: refraction) }$

$\frac{ \sin(45.0) }{ \sin(40.1) }$

= 1.26

5 0

4 years ago

Un bloque de 750 kg es empujado hacia arriba por una pista inclinada 15º respecto de la horizontal. El coeficiente de rozamiento

kotegsom [21]

Answer:

4776.98 N is the minimum force to start the rise.

Explanation:

We can use the first Newton's law to find the minimum force to move the block.

So we will have:

$F-W_{x}-F_{f}=0$

Where:

F is the force
W(x) is the weight of the block in the x direction, W = mg*sin(15)
F(f) is the static friction force (F(f) = μN), μ is the static friction coefficient 0.4.

$F=W_{x}+F_{f}=mgsin(15)+\mu N$

$F=mgsin(15)+\mu mgcos(15)$

$F=mg(sin(15)+\mu cos(15))$

$F=750*9.81(sin(15)+0.4*cos(15))$

$F=4776.98 N$

Therefore 4776.98 N is the minimum force to move the block.

I hope it helps you!

8 0

3 years ago

What role if any does air resistance have on your results

djyliett [7]

When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. ... Air resistance depends on two important factors - the speed of the object and its surface area.

6 0

3 years ago

﻿What is the usual time ice takes to evaporate above a Bunsen burner?

What is the usual time ice takes to evaporate above a Bunsen burner?