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3 years ago

If frequency is kept constant, how are the velocity and wavelength of a wave related?

2 answers:

4 0

Wave speed = (frequency) x (wavelength)

If the frequency is kept constant, then the wavelength
is directly proportional to the wave speed.

Elan Coil [88]3 years ago

3 0

<h3>Answer;</h3>

Velocity and wavelength are directly proportional when frequency is kept constant.

<h3>Explanation;</h3>

Frequency of a wave is the number of complete oscillations made by a given wave in one second.
Wavelength on the other hand, is the distance between two successful crests or troughs in a transverse wave or two successful rarefactions or compressions in a longitudinal waves.
The speed of a wave is given by the product of the frequency of a wave and the wavelength.
Speed = Frequency × wavelength, 
Therefore, if frequency is kept constant, then the speed of a wave is directly proportional to the wavelength, such that an increase in wavelength increases the speed of the wave and vice versa.

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A 5 kg block is resting on a ramp inclined at 35 degrees above the horizontal. What is the magnitude of the normal force acting

Mamont248 [21]

I'm pretty sure the answer is b 28n hope helps :)

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3 years ago

When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards

aksik [14]

Answer:

51.85m/s

Explanation:

Given parameters:

Mass of ball = 0.0459kg

Force = 2380N

Time taken = 0.001s

Unknown:

Speed of the ball afterwards = ?

Solution:

To solve this problem, we use Newton's second law of motion:

F = m x $\frac{v - u}{t}$

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

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8 0

3 years ago

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solniwko [45]

Answer:

They would land at the same time

Explanation:

They would land at the same exact time.

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3 years ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

 $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ 

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

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3 years ago

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FromTheMoon [43]

Answer:

b, a, c

Explanation:

The middle one has the shortest wavelength, then it's the top one and the last one has the longest wavelength.

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2 years ago