All categories

3 years ago

Describe a situation that demonstrates that water waved carry energy

1 answer:

5 0

<span>In the ocean in the summer time when there are high tides and you see the BIG waves, in able for those waves to get so big, they need to have some type of energy transferred on to them so that the wave may be prosees.</span><span />

You might be interested in

How does an atom develop a positive charge

Dmitry [639]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ A normal atom has the same amount of electrons and protons, making it neutral. An atom develops a positive charge when it loses an electron(s). Once it loses an electron(s), there would now be more protons that electrons.

Short answer: by losing an electron(s)

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

8 0

4 years ago

Un movil de masa 12Kg sobre el cual estan actuando varias fuerzas F_1=48N, F_2=60N y F_3=30N Calcular la aceleracion con la cual

Nikitich [7]

Answer:

Lamentablemente el problema está incompleto, pues no sabemos la dirección en la que se aplican las fuerzas. Por ello, voy a resolver el problema asumiendo dos casos. (abajo se puede ver una imagen donde se describe cada caso)

1) Todas las fuerzas están en la misma dirección.

Entonces la fuerza neta será la suma de las 3 fuerzas, entonces:

F = 48N + 60N + 30N = 138N

Y por la segunda ley de Newton sabemos que:

F = m*a

fuerza igual a masa por aceleración.

Entonces la aceleración está dada por:

a = F/m = 138N/12kg = 11.5 m/s^2

2) Segundo caso, suponemos que F1 es opuesta a F2 y F3

En este caso, la fuerza neta será:

F = F2 + F3 - F1 = 60N + 30N - 48N = 42N

En este caso, la aceleración será:

a = 42N/12kg = 3.5 m/s^2

7 0

3 years ago

A string of 26 identical Christmas tree lights are connected in series to a 120 V source. The string dissipates 73 W. What is th

spin [16.1K]

To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

$R_{eq}= \frac{V}{I}$

Here,

V = Voltage

I = Current

While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,

$P=VI \rightarrow I = \frac{P}{V}$

$I =0.608 A$

Applying Ohm's law

$R_{eq} = \frac{120V}{0.608A}$

$R_{eq} = 197.4\Omega$

Therefore the equivalent resistance of the light string is $197.4\Omega$

6 0

3 years ago

Which illustration represents low accuracy but high precision?

Misha Larkins [42]

Answer:

D i think

Explanation:

7 0

4 years ago

Read 2 more answers

The accompanying table shows measurements of the Hall voltage and corresponding magnetic field for a probe used to measure magne

aalyn [17]

0.125 mm . is the thickness of the sample.

<h3>What do you mean by hall voltage ?</h3>

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

Δ $V_{H} =\frac{I}{nqt} B$

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ $\frac{V}{T}$

b = —0.02 μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH = $100X10^{-6}$ V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

$a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26} } \\t=0.125mm$

To learn more about the hall voltage , Visit: brainly.com/question/19130911

#SPJ4

8 0

2 years ago