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Anvisha [2.4K]
3 years ago
11

Determine the stiffness of the spring to hold the 85-kg crate in the position shown.

Physics
1 answer:
GalinKa [24]3 years ago
5 0
Correct position spring hold it
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A uniform electric field is directed parallel to the +y axis. If a positive test charge begins at the origin and moves upward al
lukranit [14]

Answer: option 1 : the electric potential will decrease with an increase in y

Explanation: The electric potential (V) is related to distance (in this case y) by the formulae below

V = kq/y

Where k = 1/4πε0

Where V = electric potential,

k = electric constant = 9×10^9,

y = distance of potential relative to a reference point, ε0 = permittivity of free space

q = magnitude of electronic charge = 1.609×10^-19 c

From the formulae, we can see that q and k are constants, only potential (V) and distance (y) are variables.

We have that

V = k/y

We see the potential(V) is inversely proportional to distance (y).

This implies that an increase in distance results to a decreasing potential and a decrease in distance results to an increase in potential.

This fact makes option 1 the correct answer

3 0
3 years ago
Outside a spherically symmetric charge distribution of net charge Q, Gauss's law can be used to show that the electric field at
Sergio039 [100]

Answer:

Q at the center of the distribution.

Explanation:

  • The Gauss's law is the law that relates to the distribution of electrical charges to the resulting electrical field. It states that a flux of electricity outside the arabatory closed surface is proportional to the electricitical harg enclosed by the surface.
3 0
3 years ago
Kepler's 3rd law: harmonic law
Amanda [17]
<span>orbital velocities to their mean distances from the Sun.</span>
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3 years ago
Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and
strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance R_e defined by the formula:

\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\

and when this is combined with the third resistor in series, the equivalent resistance (R''_e) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}

The problem states that the difference between the equivalent resistances in both circuits is given by:

R''_e=R_e+630 \,\Omega

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega

8 0
3 years ago
When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink
Dovator [93]

Answer:

 y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

 Therefore the diffraction equation for slits with m = 1 remains

             a sin θ = λ

in general these experiments occur for oblique angles so

             sin θ = θ

             θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

           θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

          θ = s / R

as the angle is small the arc approaches the distance s = y

          y / R = 1.22 λ / s

          y = 1.22 λ R / a

let's calculate

            y = 1.22 500 10⁻⁹ 0.42 / 0.032

            y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

                 y <8 10⁻⁶ m

4 0
3 years ago
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