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4 years ago

What is the period of a wave with the wave speed of 12m/s and a frequency of 60 hz?

1 answer:

4 0

To determine the period of a wave you need not utilize the value for the wave speed.

The formula for the period is T= 1/f
T stands for period/time while f is for frequency. Substitute the value for the frequency which is 60 Hz. That would give you:

T= 1/60 Hz

convert Hz to cycles per second to extract the unit of time (s).

60 Hz x 1/s/1 Hz= 60/s

To continue the computation:

T=1/60/s that will give you a T= 0.017 s or 17 ms.

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Each milligram of glucose has the same amount of energy available to do work. The series B test tubes produced more bacteria per

yKpoI14uk [10]

Answer:

The series A test tube has some left amount of glucose left in it.

Explanation:

Let's assume that a fixed amount of glucose is synthesized, for the fixed quantity the bacteria produced in A and B be x and y respectively,

Therefore, the condition on x and y is, y > x as the no. of bacteria present in B is greater.

As a result B would require a greater amount of energy for its functioning, these energy would be derived from the already fixed amount of glucose present.

A test tube would also require the energy for its x number of bacteria, but it is less than that of B.

Therefore, there would be some unused glucose left in Test Tube Series A which has unused energy.

4 0

3 years ago

A lowly high diver pushes off horizontally with a speed of 2.00 m/s from the platform edge 10.0 m above the surface of the water

VladimirAG [237]

Answer:

(a) 1.6m

(b) 6.86m

Explanation:

This is a horizontal shooting problem, so we will use the following equations.

For horizontal distance x we use:

$x(t)=x_{0}+v_{0}t$

Where $x(t)$ is the horizontal distance at a time $t$ , $x_{0}$ is the initial horizontal position which in this case will be zero: $x_{0}=0$ . And $v_{0}$ is the initial speed: $v_{0}=2m/s$

So to solve (a):

(a) At what horizontal distance from the edge is the diver 0.800 s after pushing off?

we have that the time is: $t=0.8s$ so the horizontal distance is:

$x(0.8)=(2m/s)(0.8s)1.6m$

The answer for (a) is 1.6m

Now, to solve (b) we need the equation for vertical distance:

$y(t)=y_{0}-\frac{1}{2}gt^2$

Where $y(t)$ is the vertical distance at a time $t$ , $y_{0}$ is the initial vertical distance: $y_{0}=10m$ And $g$ is the gravitational acceleration: $g=9.81m/s^2$ }

(b) At what vertical distance above the surface of the water is the diver just then?

The time is the same as in the last question: $t=0.8s$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.8)^2$

$y(0.8)=10m-\frac{1}{2}(9.81m/s^2)(0.64s^2)$

$y(0.8)=10m-\frac{1}{2}(6.2784m)$

$y(0.8)=10m-(3.1392)$

$y(0.8)=6.86m$

the answer to (b) is 6.86m

(c) At what horizontal distance from the edge does the diver strike the water?

To solve (c) we first need to know how long it took to reach the height of the water, that is, for what time y = 0

Using $y(t)=y_{0}-\frac{1}{2}gt^2$

if $y(t)=0$

$0=10-\frac{1}{2}gt^2\\-10=-\frac{1}{2}gt^2\\20=gt^2\\\frac{20}{g}=t^2\\ \sqrt{\frac{20}{g}}=t$

and since $g=9.81m/s^2$

$t=\sqrt{\frac{20}{9.81} } =\sqrt{2.039}=1.43s$

At $t=1.43s$ the car hits the water, so the horizontal distance at that time is:

$x(t)=x_{0}+v_{0}t$

$x(1.43)=(2m/s)(1.43s)=2.86m$

the answer to (c) is 2.86s

6 0

3 years ago

if quasars often resemble little blue stars, what was it about them that so surprised astronomers when they were discovered?

Anna11 [10]

Quasar is famous for being an intergalactic object which is billions of years away from the earth yet can still be seen, unlike the other star body, unlike giant galaxies.

Hence, the fact that quasars can be detected from distances where even the biggest and most luminous galaxies cannot be seen means that "they must be intrinsically far more luminous than the brightest galaxies."

This condition, including other related evidence gotten in recent years concerning our galaxy, has shown that quasars are probably the central nuclei of very distant, very active galaxies.

The surprising thing was that quasars and active galaxies have a lot of mass in the center of the very small volume of the space.

Therefore, the surprising thing about quasars was that due to this mass and energy they are 100 times more luminous than Milky Way which means they have high recession velocity and a very large amount of red-shifting.

To learn more about quasars, refer: brainly.com/question/9965257

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8 0

1 year ago

Maria rides a bicycle at a velocity of 8 m sl. She brakes suddenly and stops after a distance of 2 m. What is the acceleration o

Svetach [21]

But I can help you if you need to answer this question

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8 0

3 years ago

What types of stars end their lives with supernovae?

miskamm [114]

The type of star is the north star

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3 years ago