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Sholpan [36]
2 years ago
10

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

te of 14 cm/s. How fast is the x-coordinate of the point changing at that instant?
Physics
1 answer:
blagie [28]2 years ago
7 0

Answer:7 cm/s

Explanation:

Given

Particle move along curve

y=\sqrt{1+x^3}

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

Now at (2,3)

\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}

\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s

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motikmotik

To solve this problem we will apply the concepts related to Coulomb's law for which the Electrostatic Force is defined as,

F_{initial} = \frac{kq_1q_2}{r^2}

Here,

k = Coulomb's constant

q_{1,2} = Charge at each object

r = Distance between them

As the distance is doubled so,

F_{final} = \frac{kq_1q_2}{( 2r )^2}

F_{final} = \frac{ kq_1q_2}{ 4r^2}

F_{final} = \frac{1}{4} \frac{ kq_1q_2}{r^2}

F_{final} = \frac{1}{4} F_{initial}

\frac{F_{final}}{ F_{initial}} = \frac{1}{4}

Therefore the factor is 1/4

6 0
2 years ago
If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river,
expeople1 [14]

Answer:

t=40s,

Explanation:

If you can swim in still water at 0.5m/s, the shortest time it would take you to swim from bank to bank across a 20m wide river, if the water flows downstream at a rate of 1.5m/s, is most nearly:

from the question the swimmer will have a velocity which is equal to the sum of the speed of the water and the velocity to swi across the bank

Vt=v1+v2

the time is takes to swim across the bank will be

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DY=distance across the bank

Dv=ther velocity of the swimmer across the bank

t=20/ 0.5m/s,

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velocity is the rate of displacement

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4 0
2 years ago
Read 2 more answers
Can a average velocity be zero.Why? ​
marishachu [46]

<u>Answer:</u>

Yes

<u>Explanation:</u>

Average velocity is the ratio of total displacement and time taken for that displacement:

average \space\ velocity = \frac{displacement}{time}

This means if displacement is zero, then average velocity will also be zero.

Displacement is zero when an object moves some distance in one direction, and then moves the same distance but in the opposite direction.

∴ As it is possible for displacement to be zero, it is also possible for average velocity to be zero.

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2 years ago
The Hubble Space Telescope in orbit above the Earth has a 2.4 m circular aperture. The telescope has equipment for detecting ult
strojnjashka [21]

Answer:

Option d

The minimum angular separation between two objects that the Hubble Space Telescope can resolve is 4.8x10^{-8}rad.  

Explanation:

The resulting image in a telescope that will be gotten from an object is a diffraction pattern instead of a perfect point (point spread function (PSF)).

That diffraction pattern is gotten because the light encounters different obstacles on its path inside the telescope (interacts with the walls and edges of the instrument).

 

The diffraction pattern is composed by a central disk, called Airy disk, and diffraction rings.    

 

The angular resolution is defined as the minimal separation at which two sources can be resolved one for another, or in other words, when the distance between the two diffraction pattern maxima is greater than the radius of the Airy disk.

The angular resolution can be determined in analytical way by means of the Rayleigh criterion.          

\theta = 1.22\frac{\lambda}{D}  (1)

Where \lambda is the wavelength and D is the diameter of the telescope.

Notice that it is necessary to express the wavelength in the same units than the diameter.

\lambda = 95nm \cdot \frac{1x10^{-9}m}{1nm} ⇒ 9.5x10^{-8}m

Finally, equation 1 can be used.

\theta = 1.22(\frac{9.5x10^{-8}m}{2.4m})

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Hence, the minimum angular separation between two objects that the Hubble Space Telescope can resolve is 4.8x10^{-8}rad.    

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3 years ago
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jasenka [17]

Answer:

U(3)=-43J

Explanation:

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U(0)=-3(0)^2+12(0)+C=20

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U(3)=-3*(3)^2-12(3)+20

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