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3 years ago

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 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

3 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 $10^{9}$ N $m^{2}$ / $C^{2}$

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

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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 a of

950 N = (2300 kg) a

a = (950 N) / (2300 kg)

a ≈ 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

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

v = -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]

Answer:

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 $m/s^2$

Explanation:

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