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2 years ago

If the amplitude of a wave increases by a factor of 4, how is the intensity changed?

2 answers:

8 0

The intensity increases 16 times.

The relationship between intensity and amplitude is given below:

 I = constant $A^{2}$

Since, A becomes 4 times, A' = 4A
Thus the intensity becomes, I' = constant × (4A)²

⇒ I' = 16A²

Therefore, the intensity of the wave is increased by 16 times

Kamila [148]2 years ago

6 0

the intensity of the wave is increased by 16 times

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Lostsunrise [7]

The answer is D) About 4.
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2 years ago

What is the wavelength associated with 0.113kg ball traveling with velocity of 43 m/s?

lesya [120]

Answer:

2.73×10¯³⁴ m.

Explanation:

The following data were obtained from the question:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Wavelength (λ) =?

Next, we shall determine the energy of the ball. This can be obtained as follow:

Mass (m) = 0.113 Kg

Velocity (v) = 43 m/s

Energy (E) =?

E = ½m²

E = ½ × 0.113 × 43²

E = 0.0565 × 1849

E = 104.4685 J

Next, we shall determine the frequency. This can be obtained as follow:

Energy (E) = 104.4685 J

Planck's constant (h) = 6.63×10¯³⁴ Js

Frequency (f) =?

E = hf

104.4685 = 6.63×10¯³⁴ × f

Divide both side by 6.63×10¯³⁴

f = 104.4685 / 6.63×10¯³⁴

f = 15.76×10³⁴ Hz

Finally, we shall determine the wavelength of the ball. This can be obtained as follow:

Velocity (v) = 43 m/s

Frequency (f) = 15.76×10³⁴ Hz

Wavelength (λ) =?

v = λf

43 = λ × 15.76×10³⁴

Divide both side by 15.76×10³⁴

λ = 43 / 15.76×10³⁴

λ = 2.73×10¯³⁴ m

Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.

8 0

2 years ago

A dart leaves a toy dart gun with initial velocity of 7.76 m/s, regardless of the angle it is fired. What is the maximum horizon

In-s [12.5K]

Vf=vi plus 2 ad
0=7.76 + 2(9.8)d
d=0.395m

4 0

2 years ago

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discover

givi [52]

Answer:

Acceleration due to gravity, $a=4.61\times 10^{-14}\ m/s^2$

Explanation:

It is given that,

Mass of Pluto, $m=1.4\times 10^{22}\ kg$

Distance between Neptune and Pluto, $r=4.5\times 10^{12}\ m$

The force of gravity is balanced by the gravitational force between Neptune and Pluto. It is given by :

$a=\dfrac{Gm}{r^2}$

$a=\dfrac{6.67\times 10^{-11}\times 1.4\times 10^{22}}{(4.5\times 10^{12})^2}$

$a=4.61\times 10^{-14}\ m/s^2$

So, the acceleration due to gravity at Neptune due to Pluto is $4.61\times 10^{-14}\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

2 years ago

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