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-Dominant- [34]
3 years ago
6

A trough is 10 feet long and its ends have the shape of isosceles triangles that are 3 feet across at the top and 1 foot high. I

f the trough is filled with water at a rate of 12 feet cubed per minute, how fast is the water level rising when the trough is half a foot deep?
Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
6 0

Answer:

1.6 ft/min

Explanation:

Since trough is 10 ft long and water is filled at the rate of 12ft3/min. We can calculate the rate of water filled with respect to area:

= 12 / 10 = 1.2ft2/min

As the water level rises, so does the water surface, or the bottom side of the isosceles triangles. In fact we can calculate the bottom side when the trough is half foot deep:

= 3 / 2 = 1.5 ft

The rate of change in water level would be the same as calculating the height of the isosceles triangles knowing its base

= 1.2 * 2 / 1.5 = 1.6 ft/min

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The circuit has a 3 volt EMF and two ohm resistors. How much power in watts does this circuit draw? A) 4.5 , B) 24, C) 1.13 D) 2
dlinn [17]

Answer:

P = 4.5 watts

Explanation:

Given that,

EMF of the circuit, E = 3 volt

The resistance  of the resistors, R = 2 ohms

We need to find the power of this circuit. The relation between power, emf and resistance is given by the formula as follows :

P=\dfrac{V^2}{R}

Substitute all the values,

P=\dfrac{3^2}{2}\\\\P=4.5\ W

So, the power of this circuit is equal to 4.5 watts.

5 0
3 years ago
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0
3 years ago
A rocket starting from its launch pad is subjected to a uniform acceleration of 100 meters/second2. Determine the time needed to
gizmo_the_mogwai [7]

Answer:

10s

Explanation:

Acceleration is a measure of a rate of change of velocity, or in other words, a measure of how quickly the velocity is changing.

If acceleration is constant, then the velocity is changing by a constant amount.

With an acceleration of 100 m/s^2, starting from the launching pad (and thus, an initial velocity of zero), we can calculate how long it will take to reach a final velocity of 1000m/s with the following formula:

v=at+v_o where "v" is the final velocity at some later time "t", "a" is the constant acceleration, and "v" sub-zero is the initial velocity.

v=at+v_o

(1000\text{ [m/s]})=(100 \text{ } [\text{m/s}^2] )t+(0\text{ [m/s]})

1000\text{ [m/s]}=100 \text{ } [\text{m/s}^2] *t

\dfrac{1000\text{ [m/s]}}{100 \text{ } [\text{m/s}^2]}=\dfrac{100 \text{ } [\text{m/s}^2] *t}{100 \text{ } [\text{m/s}^2]}

10\text{ [s]}=t

So, it will take 10 seconds for the rocket to reach 1000m/s when starting from the launching pad, with a constant velocity of 100m/s^2.

<u>Verification:</u>

In this situation, it is quick to verify that 10 seconds is correct by looking at what the velocities will be each second.

Recognizing that the acceleration is a=\dfrac{100 [\frac{m}{s}]}{1[s]}, the velocity increases by 100 units [m/s] every second.

At time 0[s], the velocity is 0[m/s]

At time 1[s], the velocity is 100[m/s]

At time 2[s], the velocity is 200[m/s]

At time 3[s], the velocity is 300[m/s]

At time 4[s], the velocity is 400[m/s]

At time 5[s], the velocity is 500[m/s]

At time 6[s], the velocity is 600[m/s]

At time 7[s], the velocity is 700[m/s]

At time 8[s], the velocity is 800[m/s]

At time 9[s], the velocity is 900[m/s]

At time 10[s], the velocity is 1000[m/s]

So, indeed, after 10 seconds, the velocity reaches 1000 m/s

5 0
2 years ago
Please help me with this question :
aalyn [17]

Answer:

  • 514.27 ( wavelength )

the color is green

  • 602.93 nm  ( orange color )

the observation is that there is a change of visible color

Explanation:

A) wavelength of visible light that is most strongly reflected from a point on a soap

refraction n = 1.33

wall thickness (t) = 290 nm

2nt = (2m +1 ) ∝/2 -----equation 1

note when m = 0

therefore ∝ =  4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this

when m = 1

equation 1 becomes

∝ = 4nt/3 =( 4 * 1.33 * 290) /  3 = 1542.8 / 3 = 514.27 ( wavelength )

the color is green

B) the wavelength when the wall thickness is 340 nm

∝ = 4nt / 2m +1

where m = 1

∝ = (4 * 1.33 * 340 ) / 3  = 1808.8 / 3 = 602.93 nm  ( orange color )

the observation is that there is a change of visible color

7 0
3 years ago
Which of the following has least mass
makvit [3.9K]
Answer is 1 molecule of S
4 0
3 years ago
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