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3 years ago

6. What is the wavelength of a wave that is traveling at 30 m/s and has a frequency of 3.2 Hz?

Physics

1 answer:

lidiya [134]3 years ago

6 0

Answer:

6. 9.4 m

7. 1050 m/s.

Explanation:

6. Determination of the wavelength

Velocity (v) = 30 m/s

Frequency (f) = 3.2 Hz

Wavelength (λ) =?

v = λf

30 = λ × 3.2

Divide both side by 3.2

λ = 30 / 3.2

λ = 9.4 m

Thus, the wavelength of the wave is 9.4 m.

7. Determination of the speed of the wave.

Wavelength (λ) = 350 m

Frequency (f) = 3 Hz

Velocity (v) =.?

v = λf

v = 350 × 3

v = 1050 m/s

Thus, the speed of the wave is 1050 m/s

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A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential

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Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

7 0

3 years ago

Read 2 more answers

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south. Find the magni

ozzi

Given :

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.

To Find :

The magnitude and direction of

a) A + B .

b) A - B.

Solution :

Let , direction in north is given by +j and east is given by +i .

So , $A=-63i$ and $B=63j$

Now , A + B is given by :

$A+B=-63i+63j$

$| A+B | = 63\sqrt{2}$

Direction of A+B is 45° north of west .

Also , for A-B :

$A-B=-63i-63j$

$|A-B|=63\sqrt{2}$

Direction of A-B is 45° south of west .

( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)

Hence , this is the required solution .

4 0

3 years ago

Sandi believes that people who eat at McDonald's are overweight, so she decides to do a naturalistic observation of people who e

lubasha [3.4K]

Answer:

observer bias

Explanation:

Based on the information provided within the question the thing that should concern us the most about Sandi's observations is Observer Bias. This term refers to the tendency of a researcher to see what they want as opposed to what is actually happening. This can be said because of Sandi's belief that McDonald clients are all overweight, by having this belief before actually having come to this conclusion with a series of tests, it might lead her to believe this to be true regardless of what she observes during the experiment.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

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3 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$