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3 years ago

An oscillator creates periodic waves on a stretched string.

Physics

1 answer:

sattari [20]3 years ago

6 0

Answer:

A. The wavelength doubles but the wave speed is unchanged

Explanation:

The relationship between the period and wavelength is direct. Doubling the period of the oscillator will correspondingly double the wavelength but the wave speed is unaffected

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What are facts about balloons?

Nataly_w [17]

The first rubber balloon was made by Professor Michael Faraday in 1824, out of two sheets of rubber whose edges were pressed together. Hot air balloonwas the balloon to make the first recorded manned flight. It was made by the Montgolfier brothers and launched on 21 November 1783.

3 0

3 years ago

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A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat

UNO [17]

<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

Force = Spring constant x Elongation

F = kx

Here force is weight of mass

F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x = 36 cm = 0.36 m

Substituting

F = kx

5.3 = k x 0.36

k = 14.72 N/m

Spring constant is 14.72 N/m

6 0

2 years ago

A 25 N force stretches a spring 280 cm. What was the spring constant?

boyakko [2]

Answer:

F= 25N
∆l=280cm
K=?
you apply the formula = K:F/∆l
= 0,089N/cm

4 0

2 years ago

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Which chemical equation is balanced?

nika2105 [10]

Answer:

Explanation:

<h2>Mark me me brainliest plz i really need it （⌒▽⌒）</h2>

6 0

2 years ago

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A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th

Alex73 [517]

Answer:

$V_f = 13.8863 \angle 60.89\°$

Explanation:

Our values are,

$m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s$

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

$m_1v_1=(m_1+m_2)v_fcos\theta$

For the particular case on the Y axis, we do the same with the speed of object 1.

$m_2v_2=(m_1+m_2)v_fsin\theta$

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

$Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°$

Replacing in any of the two functions, given above, we will find the final speed after the collision,

$(950)(16)=(950+1300)V_fcos(60.89)$

$V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}$

$V_f = 13.8863 \angle 60.89\°$

8 0

3 years ago