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Alexandra [31]
3 years ago
7

A wave with low energy will also have

Physics
1 answer:
Aliun [14]3 years ago
3 0
Will also have low frequency 

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An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.
bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

                        p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

                      θ = 0º

                      U = 1.16 10⁻¹⁴ 1160 cos 0

                      U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

                       θ = 180º

                      cos 180 = -1

                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

6 0
3 years ago
Hooke's law describes a certain light spring of unstretched length 33.6 cm. when one end is attached to the top of a doorframe a
masha68 [24]
Missing question: "What is the spring's constant?"

Solution:
The object of mass m=6.89 kg exerts a force on the spring equal to its weight:
F=mg=(6.89 kg)(9.81 m/s^2)=67.6 N
When the object is attached to the spring, the displacement of the spring with respect to its equilibrium position is
\Delta x=43.2 cm-33.6 cm=9.6 cm=0.096 m
And by using Hook's law, we can find the constant of the spring:
k= \frac{F}{\Delta x}= \frac{67.6 N}{0.096 m}=704.2 N/m
4 0
3 years ago
What is the force required accelerate a 5.0 kg object at 3.0m/s^ 2?
NeX [460]
The equation to find force is f=ma. So, if you plug in the information that you have you'll get F=5x3 and that'll equal F=15N
6 0
3 years ago
When Jennifer is out for a
Nady [450]
The acceleration is the principal subordinate of the speed if the speed is steady the subsidiary is invalid if the speed is diminishing the subsidiary is negative. When discussing so much stuff we consider the momentary esteem.

<span>Note that when you back off, you back off by and large yet can locally in time quicken a tiny bit, suppose amid 1/tenth of a sec since you achieved a segment of the street which was slanting. In any case, this does not change the way that when the speed diminishes, the quickening is negative.</span>
4 0
3 years ago
The magnetic field strength at the north pole of a 2.0-cmcm-diameter, 8-cmcm-long Alnico magnet is 0.10 TT. To produce the same
horrorfan [7]

Answer:

2653 turns

Explanation:

We are given that

Diameter,d=2 cm

Length of magnet,l=8 cm=8\times 10^{-2} m

1m=100 cm

Magnetic field,B=0.1 T

Current,I=2.4 A

We are given that

Magnetic field of solenoid and magnetic are same and size of both solenoid and magnetic are also same.

Length of solenoid=8\times 10^{-2} m

Magnetic field of solenoid

B=\frac{\mu_0NI}{l}

Using the formula

0.1=\frac{4\pi\times 10^{-7}\times 2.4\times N}{8\times 10^{-2}}

Where \mu_0=4\pi\times 10^{-7}

N=\frac{0.1\times 8\times 10^{-2}}{4\pi\times 10^{-7}\times 2.4}=2653 turns

6 0
3 years ago
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