What is 3,730 watts expressed as horsepwer? A. 10 B. 20 C. 5 D. 30

A double-slit experiment is set up using red light (λ = 706 nm). A first order bright fringe is seen at a given location on a sc

Elanso [62]

Answer:

λ = 470.66 nm

Explanation:

for bright fringe $y_m = \frac{m\lambda D}{d}$

D= distance between slit and screen

d= distance between the slits

for first order bright fringe m = 1,

$y_1 = \frac{1\lambda D}{d}$

$y_1 = {706*D}{d}$

for dark fringe,we have

$y_m = {(m + 1/2)\lambda D}{d}$

Now to get the dark fringes at the same location we should have;

(706)D/d = (m + 1/2)λD/d

put m = 1

(1 + 1/2)λ = (706)

λ = 470.66 nm

6 0

3 years ago

Object A is positively charged. Object A and Object B

ozzi

Answer: object B is negatively charged, object C is positively charged and object D is also positively charged

Explanation: since unlike charges attract and like charges repel, for object A which is positively charged and B to attract B must be negatively charged and then for B which is negatively charged and C to attract C must be positively charged and for C and D to repel they have to be of thesame charge which means D is positive as well.

4 0

3 years ago

The question is on the picture.

Jet001 [13]

Between noon and 2 pm, the amount of water in the rain gauge decreased.
This can be caused by evaporation, which turns water into water vapor.
Precipitation would increase the amount of rain water in the gauges, not decrease it.
Condensation occurs after evaporation but wouldn't decrease the water in the gauges by itself.
Runoff is when water on land drains into water sources such as lakes, rivers, oceans, etc.
So the answer is A. evaporation.

5 0

3 years ago

Read 2 more answers

A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th

xz_007 [3.2K]

Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

$v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}$

(b) The tension, $T$ , is given by

$v =\sqrt{\dfrac{T}{\mu}}$

where $v$ is the speed, $T$ is the tension and $\mu$ is the mass per unit length.

Hence,

$T = \mu\cdot v^{2}$

To determine $\mu$ , we need to know the mass of the cable. We use the density formula:

$\rho = \dfrac{m}{V}$

where $m$ is the mass and $V$ is the volume.

$m=\rho\cdot V$

If the length is denoted by $l$ , then

$\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}$

$T = \dfrac{\rho\cdot V}{l} v^{2}$

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

$V = \pi \dfrac{d^{2}}{4} l$

$T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2$

$T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2$

$T = 11159.4186\ldots \text{ N} = 11000 \text{ N}$

4 0

3 years ago

An alternator provides 20 A of peak current at 100 V peak AC to a resistive electric heater. How much heating power is delivered

Liono4ka [1.6K]

-- The resistance of the heater is (volts/current) = 5 ohms

-- The heating (RMS) value of a sinusoidal AC is V(peak)/√2 . For this particular alternator, V(peak)=100V, so the heating (RMS) equivalent is 70.71 V.

-- The heating power delivered to the electric heater is (E²/R).

Power = (100/√2)² / 5

Power = 5,000 / 5

Power = 1,000 watts

6 0

3 years ago