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3 years ago

A person paddles down river at an average speed of 4km

Physics

2 answers:

adoni [48]3 years ago

7 0

Nice paddling. Thanks for sharing.
Do you have some question to ask ?

almond37 [142]3 years ago

7 0

Is that a question or are you just telling us

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An air bubble at the bottom of a lake 36.0 m deep has a volume of 1.22 cm^3. If the temperature at the bottom is 5.9°C and at th

AlexFokin [52]

Answer:

volume of the bubble just before it reaches the surface is 5.71 cm³

Explanation:

given data

depth h = 36 m

volume v2 = 1.22 cm³ = 1.22 × $10^{-6}$ m³

temperature bottom t2 = 5.9°C = 278.9 K

temperature top t1 = 16.0°C = 289 K

to find out

what is the volume of the bubble just before it reaches the surface

solution

we know at top atmospheric pressure is about P1 = $10^{5}$ Pa

so pressure at bottom P2 = pressure at top + ρ×g×h

here ρ is density and h is height and g is 9.8 m/s²

pressure at bottom P2 = $10^{5}$ + 1000 × 9.8 ×36

pressure at bottom P2 =4.52 × $10^{5}$ Pa

so from gas law

$\frac{P1*V1}{t1} = \frac{P2*V2}{t2}$

here p is pressure and v is volume and t is temperature

so put here value and find v1

$\frac{10^{5}*V1}{289} = \frac{4.52*10^{5}*1.22}{278.9}$

V1 = 5.71 cm³

volume of the bubble just before it reaches the surface is 5.71 cm³

6 0

3 years ago

A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg

yulyashka [42]

Answer:

$T=245.76N$

Explanation:

We know that the frequency of the nth harmonic is given by $f_n=nf$ , where $f$ is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

$f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f$

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

$f=f_{n+1}-f_n=480Hz-400Hz=80Hz$

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

$f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}$

So the tension is:

$T=\mu(2Lf)^2$

Which for our values is:

$T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N$

6 0

3 years ago

Two astronauts push on a satellite. One pushes in the positive x direction with a force of 42 N, and the other pushes in the pos

inessss [21]

The vectors adition we can find the magnitude of the force applied by the other astronaut is 11.25 N in the y direction

Parameters given

Force of an astronaut Fₓ = 42 N

Angle θ = 15º

To find

Force another astronaut

The force is a vector magnitude for which the addition of vectors must be used, a very efficient method to perform this sum is to add the components of each vector and devise constructing the resulting vector using trigonometry and the Pythagorean theorem.

Let's use trigonometry to find the other force

tan θ = $\frac{F_y}{F_x}$