SY19-

A 150 cm long string vibrates with 3 loops and its frequency is 80 Hz. What will be the wavelength and velocity of the waves?

Vlad1618 [11]

Answer:

since each loop is ewuvivalent to one half wave lenght . the length of the string is equal to two halves of a wavelength . put in the form of an equation in the same reasoning also

5 0

2 years ago

Compare and contrast scientific theory and scientific law

grandymaker [24]

Answer:

Similarities:

Both are based on observation.
Both are supported by empirical data
Both are tested repeatedly
Both are accepted by science
Both are based on natural phenomena

Differences:

Scientific law just describes a phenomenon. It does not give an explanation as to why the phenomenon occurs. It predicts an outcome of an event based on its initial condition.
Scientific theory gives an explanation as to why an event had occurred.
Laws are more resistant to change.
Theories can be refuted and accepted, and develop other theories as well

Take note:

A scientific theory cannot be "upgraded" into a law, and a scientific law cannot be changed into a theory.

8 0

3 years ago

A plane travels 1743 KM in 2 hours 30 minutes. How fast was the plane traveling?

Advocard [28]

Answer:

$v=697.2km/h$

Explanation:

Hello.

In this case, since the velocity is computed via the division of the distance traveled by the elapsed time:

$V=\frac{d}{t}$

The distance is clearly 1743 km and the time is:

$t=2h+30min*\frac{1h}{60min} =2.5h$

Thus, the velocity turns out:

$v=\frac{1743km}{2.5h}\\ \\v=697.2km/h$

Which is a typical velocity for a plane to allow it be stable when flying.

Best regards.

5 0

2 years ago

PLS HELP 30 POINTS IF U HELP

USPshnik [31]

Answer:

its D so yeah.........

Explanation:

tell me if it's wrong

8 0

2 years ago

Read 2 more answers

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$