Ask question

Ask question

All categories

vladimir2022 [97]

2 years ago

13

An ocean wave energy device moves up and down with the motion of an ocean wave. The device produces more energy if larger waves

move the device. Which property of waves does the wave energy device use the most?
A) amplitude
B) frequency
C) volume
D) wavelength

1 answer:

Pie2 years ago

4 0

Answer:

it is D and B because they are both indicators

You might be interested in

Why did she use an red infra lamp

mixas84 [53]

To act as the Sun' was accepted but if you put 'sunlight' alone it was not accepted. The examiner wanted you to state that the infra red radiation was needed to warm up the water.

4 0

3 years ago

A Hall probe, consisting of a rectangular slab of current-carrying material, is calibrated by placing it in a known magnetic fie

Citrus2011 [14]

Answer:

(a) 0.345 T

(b) 0.389 T

Solution:

As per the question:

Hall emf, $V_{Hall} = 20\ mV = 0.02\ V$

Magnetic Field, B = 0.10 T

Hall emf, $V'_{Hall} = 69\ mV = 0.069\ V$

Now,

Drift velocity, $v_{d} = \frac{V_{Hall}}{B}$

$v_{d} = \frac{0.02}{0.10} = 0.2\ m/s$

Now, the expression for the electric field is given by:

$E_{Hall} = Bv_{d}sin\theta$ (1)

And

$E_{Hall} = V_{Hall}d$

Thus eqn (1) becomes

$V_{Hall}d = dBv_{d}sin\theta$

where

d = distance

$B = \frac{V_{Hall}}{v_{d}sin\theta}$ (2)

(a) When $\theta = 90^{\circ}$

$B = \frac{0.069}{0.2\times sin90} = 0.345\ T$

(b) When $\theta = 60^{\circ}$

$B = \frac{0.069}{0.2\times sin60} = 0.398\ T$

5 0

2 years ago

In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1

asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, $f=5\times 10^{14}\ Hz$

Separation between slits, $d=2.2\times 10^{-5}\ m$

(a) The condition for maxima is given by :

$d\ sin\theta=n\lambda$

For third maxima,

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{n\lambda}{d})$

$\theta=sin^{-1}(\dfrac{nc}{fd})$

$\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})$

$\theta=4.69^{\circ}$

(b) For second dark fringe, n = 2

$d\ sin\theta=(n+1/2)\lambda$

$\theta=sin^{-1}(\dfrac{5\lambda}{2d})$

$\theta=sin^{-1}(\dfrac{5c}{2df})$

$\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})$

$\theta=3.90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Imagine a place in the cosmos far from all gravitational and frictional influences. Suppose that you visit that place (just supp

xz_007 [3.2K]

The rock will continue to travel in a straight line with a constant velocity for ever... The reason is, once it leaves your hand there is no force acting on the rock, so it will just continue to move in a natural motion which is constant velocity.

6 0

3 years ago

A car is driving away from a crosswalk. The formula d = t 2 + 2 t expresses the car's distance from the crosswalk in feet, d , i

Ede4ka [16]

Answer:

1) No, the car does not travel at constant speed.

2) V = 9 ft/s

3) No, the car does not travel at constant speed.

4) V = 5.9 ft/s

Explanation:

In order to know if the car is traveling at constant speed we need to derive the given formula. That way we get speed as a function of time:

V(t) = 2*t + 2 Since the speed depends on time, the speed is not constant at any time.

For the average speed we evaluate the formula for t=2 and t=5:

d(2) = 8 ft and d(5) = 35 ft

$V_{2-5}=\frac{d(5)-d(2)}{5-2}=9 ft/s$

Again, for the average speed we evaluate the formula for t=1.8 and t=2.1:

d(1.8) = 6.84 ft and d(2.1) = 8.61 ft

$V_{1.8-2.1}=\frac{d(2.1)-d(1.8)}{2.1-1.8}=5.9 ft/s$

4 0

3 years ago