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guajiro [1.7K]
2 years ago
7

Can anyone help me with this question please I’ll mark as brainliest No links.

Physics
2 answers:
loris [4]2 years ago
8 0

the transverse Wave which is Answer:2

Lemur [1.5K]2 years ago
5 0

Answer:

1

Explanation:

amplitude is the maximum distance from the rest of the particles. amplitude is thus measured from the rest position to the peak of a crest or of a trough.

as seen below, graph 1 has an amplitude of 1 while both graph 2 and 3 has an amplitude of 2.

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Two drums of the same size and same height are taken.
Gelneren [198K]

Answer:

i) The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) The pressure acting on the base of <em>B</em> will be the same as the pressure acting on the base of <em>A</em>

iii) The pressure on the base of drum <em>A</em> will be slightly less than the pressure on the base of drum <em>B</em>

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

h_{max} = The height of the drums

ρ = The density of the liquid in the drum

g = The acceleration due to gravity ≈ 9.81 m/s²

i) If <em>A</em> is completely filled, we have h_A = h_{max}

Therefore, P_A = h_{max}×\rho_{liquid}×g

If <em>B</em> is half filled, we have, h_B =  (1/2)·h_{max}

P_B = (1/2) × h_{max}×\rho_{liquid}×g

Therefore, P_B = (1/2) × P_A

The pressure acting on the base of <em>B</em> will be half the pressure acting on the base of <em>A</em>

ii) If both <em>A</em> and <em>B</em> are each filled with water (the same liquid), then the pressure on their bases will be P_A = h_{max}×\rho_{water}×g = P_B, the same, given that the acceleration due to gravity, <em>g</em>, is constant and the same in Nepal and India

iii) If <em>A</em> is filled with water, and <em>B</em> is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

P_A = h_{max}×\rho_{water}×g <  P_B =

The pressure on the base of drum <em>A</em> will be less than the pressure on the base of drum <em>B.</em>

3 0
3 years ago
A 3.5 x 10-6 C charge is located 0.28 m from a 2.8 x 10-6 C charge. What is the magnitude of the force being exerted on the smal
Margarita [4]
The electrostatic force between two charges is given by Coulomb's law:
F=k_e  \frac{q_1 q_2}{r^2}
where
ke is the Coulomb's constant
q1 is the first charge
q2 is the second charge
r is the separation between the two charges

By substituting the data of the problem into the equation, we can find the magnitude of the force between the two charges:
F=(8.99 \cdot 10^{9} Nm^2C^{-2} ) \frac{(3.5 \cdot 10^{-6} N/C)(2.8 \cdot 10^{-6}N/C)}{(0.28m)^2}=1.2 N
7 0
3 years ago
The temperature of a smelting furnace is found to be 2000 degree Celsius.find the temperature on Fahrenheit scale​
Dima020 [189]

Answer:

3632.

Explanation:

Based on the formula:

2000×(9/5)+32=3632

6 0
2 years ago
The kinetic energy of an object increases as its ____ its potential energy
tankabanditka [31]
The kinetic energy of an object increases as its decreases <span>its potential energy as the sum of energy will remain constant.

In short, Your Answer would be "Decreases"

Hope this helps!</span>
8 0
3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
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