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3 years ago

HEY CAN ANYONE ANSA DIS!!!

Physics

1 answer:

Marizza181 [45]3 years ago

8 0

Answer:

C. 5.35 × 10^-5

Explanation:

5.35 × 0.00001 = 0.0000535

Hope this helped!

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"a musical tone sounded on a piano has a frequency of 261.6 hz and a wavelength of 1.31 m. what is the speed of the sound wave

Andreyy89

To solve this question, we use the wave equation which is:
C=f*λ
where:
C is the speed;
f is the frequency;
λ is the wavelength
So in this case, plugging in our values in the problem. This will give us:
C = 261.6Hz × 1.31m
= 342.696 m/s is the answer.

7 0

3 years ago

Read 2 more answers

A proton, which is the nucleus of a hydrogen atom, can be modeled as a sphere with a diameter of 2.4 fm and a mass of 1.67 10-27

Step2247 [10]

For the answer to the question above asking to d<span>etermine the density of the proton.
</span>Density is mass over volume.

The volume of a sphere is 4πr³/3. r is half the diameter.

So the density would be 2.3×10¹⁷ kg/m³.
I hope my answer helped you. Feel free to ask more questions. Have a nice day!

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4 years ago

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1 A ball is rolling along the floor with a constant velocity of 8 m/s. How far will it have gone after

Dennis_Churaev [7]

Answer:

448 meters

Explanation:

every second it moves 8 meters, so all you have to do is multiply 56x8 or 8x56 either way it is the same thing and you will get the same answer

3 0

3 years ago

Principal of simple machine?

sergij07 [2.7K]

Answer:

if there is no friction in a simple machine, work output and work input are found equal in that machine

Explanation:

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3 years ago

1. A 2.5 kg led projector is launched as a projectile off a tall building. At one point, as it

spin [16.1K]

Answer:

Explanation:

I got everything but i. Don't know why but it's eluding me. So let's do everything but that.

a. PE = mgh so

PE = (2.5)(98)(14) and

PE = 340 J

b. $KE=\frac{1}{2}mv^2$ so

$KE=\frac{1}{2}(2.5)(14)^2$ and

KE = 250 J

c. TE = KE + PE so

TE = 340 + 250 and

TE = 590 J

d. PE at 8.7 m:

PE = (2.5)(9.8)(8.7) and

PE = 210 J

e. The KE at the same height:

TE = KE + PE and

590 = KE + 210 so

KE = 380 J

f. The velocity at that height:

$380=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(380)}{2.5} }$ so

v = 17 m/s

g. The velocity at a height of 11.6 m (these get a bit more involed as we move forward!). First we need to find the PE at that height and then use it in the TE equation to solve for KE, then use the value for KE in the KE equation to solve for velocity:

590 = KE + PE and

PE = (2.5)(9.8)(11.6) so

PE = 280 then

590 = KE + 280 so

KE = 310 then

$310=\frac{1}{2}(2.5)v^2$ and

$v=\sqrt{\frac{2(310)}{2.5} }$ so

v = 16 m/s

h. This one is a one-dimensional problem not using the TE. This one uses parabolic motion equations. We know that the initial velocity of this object was 0 since it started from the launcher. That allows us to find the time at which the object was at a velocity of 26 m/s. Let's do that first:

$v=v_0+at$ and

26 = 0 + 9.8t and

26 = 9.8t so the time at 26 m/s is

t = 2.7 seconds. Now we use that in the equation for displacement:

Δx = $v_0t+\frac{1}{2}at^2$ and filling in the time the object was at 26 m/s:

Δx = 0t + $\frac{1}{2}(-9.8)2.7)^2$ so

Δx = 36 m

i. ??? In order to find the velocity at which the object hits the ground we would need to know the initial height so we could find the time it takes to hit the ground, and then from there, sub all that in to find final velocity. In my estimations, we have 2 unknowns and I can't seem to see my way around that connundrum.

4 0

3 years ago