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3 years ago

In which medium would sound travel the fastest

Physics

2 answers:

mezya [45]3 years ago

7 0

Answer:D

Explanation:

Fofino [41]3 years ago

6 0

Answer:

It think it's either A or D.

But I think the answer is most likely to be A because most railroad tracks aren't in tunnels and the D doesn't describe anything about tunnels.

Explanation:

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A car traveling at 31 m/s starts to decelerate steadily. It comes to a complete stop in 7 seconds. What is its acceleration

N76 [4]

Vf=V1+at
0=31+a(7)
-31/7=a
a=-4.43m/s^2

Hope this helps.

8 0

3 years ago

What is the half-life of isotope X?

Brrunno [24]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

A fluid moves through a tube of length 1 meter and radius r=0.002±0.0002 meters under a pressure p=4⋅105±1750 pascals, at a rate

yaroslaw [1]

Answer:

The maximum error is $\Delta \eta = 2032.9$

Explanation:

From the question we are told that

The length is $l = 1\ m$

The radius is $r = 0.002 \pm 0.0002 \ m$

The pressure is $P = 4 *10^{5} \ \pm 1750$

The rate is $v = 0.5*10^{-9} \ m^3 /t$

The viscosity is $\eta = \frac{\pi}{8} * \frac{P * r^4}{v}$

The error in the viscosity is mathematically represented as

$\Delta \eta = | \frac{\delta \eta}{\delta P}| * \Delta P + |\frac{\delta \eta}{\delta r} |* \Delta r + |\frac{\delta \eta}{\delta v} |* \Delta v$

Where $\frac{\delta \eta }{\delta P} = \frac{\pi}{8} * \frac{r^4}{v}$

and $\frac{\delta \eta }{\delta r} = \frac{\pi}{8} * \frac{4* Pr^3}{v}$

and $\frac{\delta \eta }{\delta v} = - \frac{\pi}{8} * \frac{Pr^4}{v^2}$

$\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]$

substituting values

$\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]$

$\Delta \eta = \frac{\pi}{8} [56 + 5120 ]$

$\Delta \eta = 647 \pi$

$\Delta \eta = 2032.9$

4 0

4 years ago

A 22.0 kg child plays on a swing having support ropes that are 1.90 m long. A friend pulls her back until the ropes are 45.0 ∘ f

love history [14]

Answer:

120 J

Explanation:

Given:

Mass of the child (m) = 22.0 kg

Length of the rope (L) = 1.90 m

Angle made by the rope (x) = 45.0°

Consider the diagram below for the given scenario.

Point A represents the bottom position, point B represents the maximum height reached, OA or OB is the length of rope.

OA = OB = L = 1.90 m

∠ AOB = x = 45.0°

Now, the maximum height reached by the child is given as:

$h=L-L\cos x\\\\h=L(1-\cos x)\\\\h=1.90\ m(1-\cos (45.0))\\\\h=0.556\ m$

So, as the child is pulled to a height 'h', the child's potential energy is increased.

Therefore, the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing is equal to the increase in the gravitational potential energy of the child.

The increase in gravitational potential energy is given as:

$\Delta U=mgh\\\\\Delta U=(22.0\ kg)(9.8\ m/s^2)(0.556\ m)\\\\\Delta U=119.9\approx 120\ J$

Therefore, the potential energy for the child just as she is released, compared with the potential energy at the bottom of the swing is 120 J.

5 0

3 years ago

The kinetic energy of an object increases as its ____ its potential energy

tankabanditka [31]

The kinetic energy of an object increases as its decreases <span>its potential energy as the sum of energy will remain constant.

In short, Your Answer would be "Decreases"

Hope this helps!</span>

8 0

3 years ago