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2 years ago

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In science, Bob learns that the energy of a wave is directly proportional to the square of the waves amplitude. If the energy of

a wave is 4 J, what should he predict as the amplitude of the wave?

1 answer:

KIM [24]2 years ago

4 0

The energy of a wave is directly proportional to the square of the waves amplitude. Therefore, E = A² where A is the amplitude. This therefore means when the amplitude of a wave is doubled the energy will be quadrupled, when the amplitude is tripled the energy increases by a nine fold and so on.
Thus, in this case if the energy is 4J, then the amplitude will be √4 = 2 .

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ASAPP PLS HELP MEE

Anastasy [175]

Answer:

B) 2.7 g of aluminium has a volume of 1 cm^3

Explanation:

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the equation;

$Density = \frac{mass}{volume}$

If the density of aluminum is 2.7 g/cm³, it simply means that 2.7 g of aluminium has a volume of 1 cm³

Check:

Given the following data;

Mass = 2.7 grams

Volume = 1 cm³

Substituting into the formula, we have;

$Density = \frac{2.7}{1}$

Density = 2.7 g/cm³

7 0

1 year ago

Which section of the reaction represents the reactants? A) a B) b C) c D) d

ivann1987 [24]

This question is incomplete, but I can do it for you, considering the equation to be *In its most famous form*:
A+B⇒C+D
A and B here are the reactants, while C and D are the products.
The reactants are generally the input materials in the beginning of any chemical reactions and they usually, if not always, are on the left hand side of the chemical equation. While the products are on the right hand side and are the final output of the chemical reaction.

Hope this helps.

3 0

2 years ago

Read 2 more answers

A charge +1.9 μC is placed at the center of the hollow spherical conductor with the inner radius 3.8 cm and outer radius 5.6 cm.

Archy [21]

To solve this problem we will apply the concepts related to load balancing. We will begin by defining what charges are acting inside and which charges are placed outside.

PART A)

The charge of the conducting shell is distributed only on its external surface. The point charge induces a negative charge on the inner surface of the conducting shell:

$Q_{int}=-Q1=-1.9*10^{-6} C$ . This is the total charge on the inner surface of the conducting shell.

PART B)

The positive charge (of the same value) on the external surface of the conducting shell is:

$Q_{ext}=+Q_1=1.9*10^{-6} C$

The driver's net load is distributed through its outer surface. When inducing the new load, the total external load will be given by,

$Q_{ext, Total}=Q_2+Q_{ext}$

$Q_{ext, Total}=1.9+3.8$

$Q_{ext, Total}=5.7 \mu C$

5 0

3 years ago

The speed of an electromagnetic wave traveling in a transparent nonmagnetic substance is v = 1/ κμ0ε0 , where κ is the dielectri

vladimir1956 [14]

Answer:

$v=2.24\times 10^{8}\ m/s$

Explanation:

Given that,

The speed of an electromagnetic wave traveling in a transparent nonmagnetic substance is given by :

$v=\dfrac{1}{\sqrt{k\mu_o\epsilon_o}}$

Where

k is the dielectric constant of the substance.

v is the speed of light in water

$c=\dfrac{1}{\sqrt{1.78\times 4\pi \times 10^{-7}\times 8.85\times 10^{-12}}}$

$v=2.24\times 10^{8}\ m/s$

So, the speed of light in water is $2.24\times 10^{8}\ m/s$

7 0

2 years ago

A square loop of side 7 cm is placed with the nearest side 2 cm from a long wire carrying a current that varies with time at a c

ioda

Answer:

Explanation:

side of the square loop, a = 7 cm

distance of the nearest side from long wire, r = 2 cm = 0.02 m

di/dt = 9 A/s

Integrate on both the sides

$\int _{0}^{i}di =9\int _{0}^{t}dt$

i = 9t

(a) The magnetic field due to the current carrying wire at a distance r is given by

$B = \frac{\mu_{0}i}{2\pi r}$

$B = \frac{\mu_{0}\times 9t}{2\pi r}$

(b)

Magnetic flux,

$\phi=\int B\times a dr$

$\phi=\int \frac{\mu_{0}\times 9t}{2\pi r}\times a dr$

$\phi=\frac{\mu_{0}\times 9t\times a}{2\pi}\times ln\left ( \frac{2 + 7}{2} \right )$

$\phi=\frac{\mu_{0}\times 9t\times 0.07}{2\pi}\times ln(4.5)$

$\phi = 1.89 \times 10^{-7}t$

(c)

R = 3 ohm

$e = -\frac{d\phi}{dt}$

magnitude of voltage is

e = 1.89 x 10^-7 V

induced current, i = e / R = (1.89 x 10^-7) / 3

i = 6.3 x 10^-8 A

8 0

2 years ago