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Dafna1 [17]
4 years ago
15

An electric field of magnitude e is measured at a distance r from a point charge q. if the charge is doubled to 2q, and the elec

tric field is now measured at a distance of 2r from the charge, what is the value of the new electric field?
Physics
1 answer:
pav-90 [236]4 years ago
5 0
Electric field is proportional to the charge and inversely proportional to the square of distance:
E\propto\frac{Q}{r^{2}}
With charge 2q and distance 2r, the electric field is proportional to:
E_{q}\propto\frac{2q}{(2r)^{2}}=\frac{2q}{4r^{2}}=\frac{1}{2}\frac{q}{r^{2}}
The new electric field is half of that of the original measured field.
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Please help I don't know how to answer these questions!
Yuki888 [10]

1) The potential energy is the most at the highest position and the least at the equilibrium position

2) The kinetic energy is the most at the equilibrium position and  the least at the highest position

Explanation:

1)

The potential energy of an object is the energy possessed by the object due to its position in a gravitational field; mathematically, it is given by

PE=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the pendulum in this problem, m is the mass of the bob, and h is the height of the above relative to the ground. We see from the formula that the potential energy is directly proportional to the height:

PE\propto h

This means that:

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  • The potential energy is the least when the bob is at the equilibrium position,  which is the lowest position

2)

We can solve this part by applying the law of conservation of energy: in fact, the total mechanical energy of the pendulum (sum of potential and kinetic energy) is constant at any time during the motion,

E=KE+PE=const.

where KE is the kinetic energy.

From the equation above, we observe that:

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Therefore, this implies that:

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3 years ago
The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a
grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

\sin\theta=\dfrac{3}{5}

The horizontal component is

T_{x}=T\cos\theta

The vertical component is

T_{y}=T\sin\theta

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'

Put the value into the formula

0=190\times2+300\times 4-T\sin\theta\times 4

T\sin\theta\times 4=380+1200

T=\dfrac{1580\times5}{3\times 4}

T=658.33\ N

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

F_{x}=T\cos\theta

Put the value into the formula

F_{x}=658.33\times\dfrac{4}{5}

F_{x}=526.66\ N

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

F_{y}=F_{b}+F_{w}-T\sin\theta

Put the value into the formula

F_{y}=190+300-658.33\times\dfrac{3}{5}

F_{y}=95.002\ N

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(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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