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Nina [5.8K]
2 years ago
5

A cowboy at a dude ranch fills a horse trough that is 1.53 m long, 61 cm wide, and 42 cm deep. He uses a 2.0 cm diameter hose fr

om which water emerges at 1.25 m/s. How long does it take him to fill the trough? Caution: 2.0 cm is the diameter of the hose, not the radius.
Physics
1 answer:
Hatshy [7]2 years ago
7 0

Answer:

Length of the trough is

l

=

1.5

m

The breadth of the trough is

w

=

60

c

m

Height of the trough is

h

=

40

c

m

The volume of the trough is

V

=

l

×

b

×

h

V

=

(

1.5

)

×

(

0.6

)

×

(

0.4

)

V

=

0.36

m

3

The diameter of the hose pipe is

A

=

π

d

2

4

A

=

(

π

)

(

0.02

2

4

)

A

=

3.14

×

10

−

4

m

2

Now for the rate of flow:

Q

=

A

×

V

Q

=

(

3.14

×

10

−

4

)

×

(

0.36

)

Q

=

1.13

×

10

−

4

m

3

Now for the time:

t

=

V

Q

t

=

0.36

1.13

×

10

−

4

t

=

3185.8

s

Thus, the time taken in order to fill the tank is 3185.8 s

Explanation:

this is the answer

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Two conducting spheres, one having twice the diameter of the other, are separated by a distance that is large compared to their
Snezhnost [94]

Answer: Option (a) is the correct answer.

Explanation:

When these two conducting spheres are connected together through a thin wire then charge from the smaller sphere will travel through the wire. And, this charge will continue to travel towards the neutral sphere until the charge on both the spheres will become equal to each other.

For example, charge on small sphere is 5 C then this charge will continue to travel towards the neutral sphere until its charge also becomes equal to 5 C.

Hence, then their potential will also become equal.

Thus, we can conclude that the spheres are connected by a long, thin wire, then after a sufficiently long time the two spheres are at the same potential.

6 0
2 years ago
Bianca is standing at x =600m. Firecracker 1, at the origin, and firecracker 2, at x =900m, explode simultaneously. The flash fr
agasfer [191]

Answer:3 \mu s

Explanation:

Given

Bianca is at x=600 m

i.e. distance between origin and Bianca is 600 m

time taken to reach Bianca eyes is

t=\frac{600}{speed\ of\ light}

t=\frac{600}{3\times 10^8}

t=2\times 10^{-6} s

t=2 \mu s

i.e. Cracker exploded at t=2\mu s because it is observed at t=4\mu s

Time taken by second cracker flash to reach Bianca eyes

t_2=\frac{300}{3\times 10^8}

t_2=10^{-6}

t_2=1 \mu s

Therefore it will be observed at t=3 \mu s

4 0
2 years ago
Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming
Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

tan \ \theta = \dfrac{12}{100} \\ \\  tan \ \theta = 0.12 \\ \\  \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0

The drag force D = Wsinθ

\dfrac{1}{2}C_v \rho AV^2 = W sin \theta

where;

\rho = 1.23 \ kg/m^3

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}

C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}

C_D =0.9378

8 0
2 years ago
Reset the PhET simulation (using the button in the lower right) and set it up in the following manner: select Oscillate, select
lubasha [3.4K]

Answer:

Check Explanation

Explanation:

This is a question that is as a result of an experimental procedure.

The Phet simulation is put on the settings given in the question;

- select Oscillate

- Select No End

- Use the parameters in parentheses by sliding the bars for Amplitude (1.00 cm), Frequency (1.40 Hz)

- Damping (none)

- Tension (highest)

I'll attach an interface of the Phet simulation to this solution.

Once all of these settings have been fixed, the simulation gives a wave pattern whose wavelength can be read from the ruler attached to the background of the simulation.

The wavelength is the distance from crest to crest or from trough to trough.

From the simulation example I have attached, the distance from crest to crest is from the green indicator on one crest to the green indicator on the next crest, that is about 5 to 5.1 cm

The velocity of a wave, v, is related to the frequency, f, and wavelength, λ of the wave through

v = fλ

For the photon, the velocity of the wave is the speed of light,

v = c = (3.00 × 10⁸) m/s

The wavelength computed from the simulation = λ = 5.1 cm = 0.051 m

c = fλ

frequency = (c/λ) = (3.00 × 10⁸) ÷ 0.051 = (5.88 × 10⁹) Hz

So, this step can be used to obtaim rhe required frequency of the photon, just follow these steps and use the calculation method too. You should be able to obtain the frequency of the photon in your experiment.

Hope this Helps!!!

5 0
2 years ago
Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la
rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93  ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

v=v_0-g \,t

for the object's vertical displacement we have

y-y_0=v_0\,t - \frac{g}{2} \,t^2

If the maximum height reached by the object is given in meters, we use the value for g in m/s^2 which is: 9.8\,\,m/s^2

If the maximum height of the object is given in feet, we use the value for g in  ft/s^2  which is : 32\,\,ft/s^2

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}

and now we use this to express the maximum height in the second equation we typed:

y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g})  - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}

Then if the max height is 1.1 meters, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s

If the max height is 1.1 feet, we use the following formula to solve for v_0:

1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s

5 0
2 years ago
Read 2 more answers
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