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cestrela7 [59]
3 years ago
11

HELP ;

Physics
2 answers:
mestny [16]3 years ago
7 0

Answer:

The frequency would double.

Explanation:

Given:

Speed of wave (v) = constant.

Frequency of wave initially (f₁) = 2 Hz

Initial wavelength of the wave (λ₁) = 1 m

Final wavelength of the wave (λ₂) = 0.5 m

Final frequency of the wave (f₂) = ?

We know that the product of wavelength and frequency of the wave is equal to the speed of the wave.

Therefore, framing in equation form, we have:

Wavelength × Frequency = Speed

\lambda\times f=v

It is given that speed of the wave remains the same. So, the product must always be a constant.

Therefore,

\lambda\times f=constant\ or\ \\\lambda_1\times f_1=\lambda_2\times f_2

Now, plug in the given values and solve for 'f₂'. This gives,

1\times 2=0.5\times f_2\\\\f_2=\frac{2}{0.5}=4\ Hz

Therefore, the final frequency is 4 Hz which is double of the initial frequency.

f₂ = 2f₁ = 2 × 2 = 4 Hz

So, the second option is correct.

Veronika [31]3 years ago
7 0

Answer:

The answer is C The frequency would double.

Explanation:

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F(x)= 10x-5<br>What is the value of f-1(-4) ?​
Afina-wow [57]

Answer:

f^{-1}(-4) = \frac{1}{10}

Explanation:

Firstly finding f^{-1}(x)

So,

f(x) = 10x-5

Substitute y = f(x)

y = 10x-5

Exchange the values of x and y

x = 10y-5

Solving for y

x = 10y-5

Adding 5 to both sides

10y = x+5

Dividing both sides by 10

y = \frac{x+5}{10}

Replace y = f^{-1}(x)

f^{-1}(x) = \frac{x+5}{10}

For x = -4

f^{-1}(-4) = \frac{-4+5}{10}

f^{-1}(-4) = \frac{1}{10}

6 0
2 years ago
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Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend
Rasek [7]

Answer:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m

Explanation:

Alpha particles Ichargeq q = + 2e mass m=6.8*10^ -27 kg) at 17*10^ 4 m/s What magnetic field strength would be required to bend them into a circular path of radiuse c = 0.25m ok

3 0
3 years ago
Find the direction and magnitude of the net force exerted on the point charge q3 in the figure. Let q= +2.4 μC and d= 33cm.
kobusy [5.1K]

With the use of electric force formula, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees

ELECTRIC FORCE (F)

F = \frac{KQq}{d^{2} }

Where K = 9 x 10^{9} Nm^{2}/C^{2}

The distance between q_{1} and q_{3} can be calculated by using Pythagoras theorem.

d = \sqrt{33^{2} + 33^{2}  }

d = 46.7 cm = 0.467 m

For force F_{1}, substitute all the parameters into the formula above

F_{1} = (9 x 10^{9} x 3 x 1)/0.467^{2}

F_{1} = 2.7 x 10^{10}/0.218

F_{1} = 1.24 x 10^{11} N

For force F_{4}, substitute all the parameters into the formula above

F_{4} = (9 x 10^{9} x 3 x 4)/0.33^{2}

F_{4} = 1.08 x 10^{11}/0.1089

F_{4} = 9.92 x 10^{11} N

For force F_{2}, substitute all the parameters into the formula above

F_{2} = (9 x 10^{9} x 3 x 2)/0.33^{2}

F_{2} = 5.4 x 10^{10}/0.1089

F_{2} = 4.96 x 10^{11} N

Summation of forces on Y component will be

F_{y} = F_{4} - F_{1} Sin 45

F_{y} = 9.92 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{y} = 9.04 x 10^{11} N

Summation of forces on X component will be

F_{x} = F_{2} - F_{1} Cos 45

F_{x} = 4.96 x 10^{11} - 1.24 x 10^{11} Sin 45

F_{x} = 4.08 x 10^{11} N

Net Force = \sqrt{F_{x} ^{2} + F_{y} ^{2}  } }

Net force = \sqrt{(4.08*10^{11}) ^{2} + (9.04*10^{11}) ^{2}  }

Net force = 9.9 x 10^{11} N

The direction will be

Tan ∅ = F_{y}/F_{x}

Tan ∅ = 9.04 x 10^{11} / 4.08 x 10^{11}

Tan ∅ = 2.216

∅ = Tan^{-1}(2.216)

∅ = 65.7 degrees

Therefore, the direction and magnitude of the net force exerted on the point charge q3 are 9.9 x 10^{11} N and 66 degrees approximately.

Learn more about electric Force here: brainly.com/question/4053816

8 0
2 years ago
Read 2 more answers
A 2300 kg sailboat is moving west at 5.5 m/s when an eastward wind
pogonyaev

The boat is initially at equilibrium since it seems to start off at a constant speed of 5.5 m/s. If the wind applies a force of 950 N, then it is applying an acceleration <em>a</em> of

950 N = (2300 kg) <em>a</em>

<em>a</em> = (950 N) / (2300 kg)

<em>a</em> ≈ 0.413 m/s²

Take east to be positive and west to be negative, so that the boat has an initial velocity of -5.5 m/s. Then after 11.5 s, the boat will attain a velocity of

<em>v</em> = -5.5 m/s + <em>a</em> (11.5 s)

<em>v</em> = -0.75 m/s

which means the wind slows the boat down to a velocity of 0.75 m/s westward.

5 0
3 years ago
PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!
SVETLANKA909090 [29]

<u>Answer:</u>

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 m/s^2

<u>Explanation:</u>

1)  Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2)  Average speed = Total distance covered / Time taken for that distance to cover.

    Total distance covered = 2+0.5+2.5 = 5 km

    Time taken = 3 hours

     Average speed = 5/3 = 1.66 km /hour

3)    Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

  Change in velocity = 14 - 6 = 8 m/s

   Time = 4 seconds

   So acceleration = 8 / 4 = 2 m/s^2

6 0
3 years ago
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