Which of the following is a likely outcome of a failed experiment?

A blow-dryer and a vacuum cleaner each operate with a voltage of 120 V. The current rating of the blow-dryer is 13 A, while that

Feliz [49]

Answer:

(a) 1560 W

(b) 576 W

(c) 1.01

Explanation:

Voltage, V = 120 V

Current in dryer, I = 13 A

current in vacuum cleaner, i' = 4.8 A

(a) Power consumed by dryer,

P = V I = 120 x 13 = 1560 W

(b) Power consumed by vacuum cleaner

P' = V I' = 120 x 4.8 = 576 W

(c) Energy consumed by dryer

E = P x t = 1560 x 15 x 60 = 1404000 J

Energy consumed by the vacuum cleaner

E' = P' x t' = 576 x 40 x 60 = 1382400 J

the ratio of energies is

E : E' = 1404000 : 1382400 = 1.01

3 0

3 years ago

What defines minerals will mark brainlest if croccted right

zvonat [6]

<span> Uniform chemical composition
Solid
Crystalline structure </span>these are the characteristics of minerals

3 0

3 years ago

Why does the total amount of energy before and after<br> any energy transformations remain the same?

snow_tiger [21]

The Law of Conservation of Energy states that, in an isolated system, energy remains constant and can not be created or destroyed, only transferred from one form to another. This law was created by Julius Robert Mayer.

8 0

3 years ago

Read 2 more answers

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

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