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USPshnik [31]
3 years ago
11

(HELP!!! 30 pts if answered right. )What formula gives the strength of an electric field, E, at a distance from a known source c

harge? E = Startfraction F Subscript e baseline over q times d Endfraction. E = Startfraction k times q over d Endfraction. E = Startfraction k times q over d Superscript 2 baseline Endfraction. E = Startfraction F Subscript e baseline over d Endfraction.
Physics
1 answer:
umka2103 [35]3 years ago
8 0

Answer:

E=\frac{k\,Q}{d^2}

Explanation:

The strength of an electric field E produced by a single charge Q at a distance d from it is given by the formula: E=\frac{k\,Q}{d^2}, where K represents the Coulomb constant.

Since the electric field E is derived from the Coulomb Force per unit charge using a positive test charge, the field's units will be in units of Newtons/Coulomb, and be the formula for the Coulomb electric force between to charges (Q1 and Q2),

F_C=k\frac{Q_1\,Q_2}{d^2}

but modified with only one charge showing in the numerator of the expression.

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If the spring constant is doubled, what value does the period have for a mass on a spring? A. The period would double by square
antoniya [11.8K]

Answer:

The period would decrease by sqrt(2)

Explanation:

The restoring force is given by,

F = -kx

According to Newton's second law of motion,

ma = -kx

ma + kx = 0

The time period is given by,

T =\frac{2\pi }{\omega }

Where \omega is the angular velocity and it is given by,

\omega = \sqrt{\frac{k}{m} }

Now if the spring constant is doubled then,

k_{2} = 2k

Thus,

T_{2} =\frac{2\pi }{\sqrt{\frac{2k}{m} } }

\frac{T_{2} }{T} = \frac{\frac{2\pi }{\sqrt{\frac{2k}{m} } }}{\frac{2\pi }{\sqrt{\frac{k}{m} } }}

\frac{T_{2} }{T} = \sqrt{\frac{k}{2k} } = \sqrt{\frac{1}{2} }

T_{2} = \frac{T}{\sqrt{2} }

Thus, The period would decrease by sqrt(2).

Hence, option D is correct.

3 0
2 years ago
Read 2 more answers
A 975-kg elevator accelerates upward at 0.754 m/s2, pulled by a cable of negligible mass. Find the tension force in the cable.
Zina [86]

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

\sum F = ma

T-mg = ma

Here,

T = Tension

m = Mass

g = Gravitational Acceleration

a = Acceleration (upward)

Rearranging to find T,

T = m(g+a)

T = (975)(9.8+0.754)

T= 10290.15N

Therefore the tension force in the cable is 10290.15N

7 0
3 years ago
Apply the impulse-momentum relation and the work-energy theorem to calculate the maximum value of t if the cake is not to end up
loris [4]
Puto chupame el semen ok? right?
8 0
3 years ago
A 221 g cart starts from rest and rolls in the right direction (positive) down an incline. The incline is at a height of 5 cm. A
Julli [10]

Answer:

1) p₀ = 0.219 kg m / s, p = 0, 2)  Δp = -0.219 kg m / s, 3) 100%

Explanation:

For the first part, which is speed just before the crash, we can use energy conservation

Initial. Highest point

            Em₀ = U = mg y

Final. Low point just before the crash

           Emf = K = ½ m v²

          Em₀ = Emf

          m g y = ½ m v²

           v = √ 2 g y

Let's calculate

           v = √ (2 9.8 0.05)

           v = 0.99 m / s

1) the moment before the crash is

           p₀ = m v

           p₀ = 0.221 0.99

           p₀ = 0.219 kg m / s

After the collision, the car's speed is zero, so its moment is zero.

           p = 0

2) change of momentum

           Δp = p - p₀

            Δp = 0- 0.219

            Δp = -0.219 kg m / s

3) the reason is

     Δp / p = 1

In percentage form it is 100%

3 0
3 years ago
A football is thrown horizontally with an initial velocity of(16.6 {\rm m/s} ){\hat x}. Ignoring air resistance, the average acc
Ray Of Light [21]

Answer:

A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s  c) 46º below horizontal.

Explanation:

A) Once released, the football is not under the influence of any external force in the horizontal direction, so it  continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.

If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that  this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:

vₓ = 16.6 m/s i

In the vertical direction, the football, once released, is in free fall, starting from rest.

So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:

vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s

Assuming that the upward direction is the positive  for the y-axis (perpendicular to the chosen  x-axis), we can write the vertical component of  the velocity vector, at t=1.75 s, as follows:

vy = -17.2 m/s j

So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:

v = 16.6 m/s i -17.2 m/s j

b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:

v =\sqrt{(16.6m/s)^{2}+ (-17.2m/s)^{2}} = 23.9 m/s

v = 23.9 m/s

c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:

tg θ =\frac{-17.2}{16.6} =-1.036

⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.

6 0
3 years ago
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