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2 years ago

Hii! help asap. i’ll give brainliest thanks!

Physics

1 answer:

o-na [289]2 years ago

5 0

I believe it’s the mass of the box but I don’t no if I’m right

Hope this helped

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What kind of energy is stored when you squeeze a mattress?

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1.is anyone else bored with school<br><br> 2. is anyone getting out of school soon or at 3

storchak [24]

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i am bored and i get out in 10 minuets

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2 years ago

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

Lemur [1.5K]

Answer:

a) v = 0.9167 m / s, b) A = 0.350 m, c) v = 0.9167 m / s, d) A = 0.250 m

Explanation:

a) to find the velocity of the wave let us use the relation

v = λ f

the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength

λ = x

the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period

T / 2 = t

T = 2t

period and frequency are related

f = 1 / T

f = 1 / 2t

we substitute

v = x / 2t

v = 5.50 / 2 3

v = 0.9167 m / s

b) the amplitude is the distance from a maximum to zero

2A = y

A = y / 2

A = 0.700 / 2

A = 0.350 m

c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same

v = 0.9167 m / s

d) the amplitude is

A = 0.500 / 2

A = 0.250 m

4 0

2 years ago

Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago