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natita [175]
3 years ago
5

To review relationships among electric potential, electric potential energy, and force on a test charge

Physics
1 answer:
adelina 88 [10]3 years ago
5 0

Answer:

positive, negative, positive

Explanation:

Electric field lines are the imaginary lines that is the path followed by an isolated unit positive charged particle in an electric field.

The direction of electric field at that point is determined by drawing the tangent at that point on the electric field line.

Part A

The electric field lines always begin at <u>positive</u> charges and end at <u>negative</u> charges.

One could also say that the lines we use to represent an electric field indicate the direction in which a <u>positive</u> test charge would initially move when released from rest.

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Cellular phones use _____?
dangina [55]
Let's see: frequency of cellular phone waves (GSM phones) is (800-1900 MHz). If we look at the table of the electromagnetic spectrum, we can see that this range is contained within the frequencies of the microwaves, which include waves in the range 300 MHz-300 GHz.

So, summarizing, the correct answer is "microwaves".
3 0
3 years ago
Read 2 more answers
A neutron has
zavuch27 [327]

Answer:

D

Explanation:

a neutron does not have a positive nor negative charge it remains neutral

6 0
2 years ago
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The 10-lb block has a speed of 4 ft/s when the force of f=(8t2)f=(8t2) lb is applied. determine the velocity of the block when t
KatRina [158]

The velocity of the block when t == 2 s is 60.7 ft./sec.

Equations of Motion.

Here the friction is F_f = \mu_k N = 0.2 N

+ \uparrow \sum F_y = ma_y; \quad N – 10 = \frac { 10 } { 32.2 }(0) \quad N = 10 lb \\ \begin{aligned} \underrightarrow{ + } \sum F_x = ma_x; \quad 8t^2 – 0.2(10 &) = \frac { 10 } { 32.2 }a \\ & a = 3.22(8t^2 – 2) ft/s^2 \end{aligned}

Kinematics.

The velocity of the block as a function of t can be determined by

integrating dv = adt using the initial condition v = 4 ft./s at t = 0.

\int_{ 4 ft/s }^{ v } dv = \int_0^t 3.22(8t^2 – 2)dt \\ \begin{aligned} v – &4 = 3.22 (\frac 8 3 t^3 – 2t) \\ & v = \{8.5867t^3 – 6.44t + 4 \} ft/s \end{aligned}

The displacement as a function of t can be determined by integrating

ds = vdt using

the initial condition s = 0 at t = 0

\int_0^s ds = \int_0^t (8.5867t^3 – 6.44t + 4)dt \\ s = \{2.1467t^4 – 3.22t^2 + 4t \} ft

at t = 2 sec

s = 30 ft.

Thus, at s = 30 ft.,

\begin{aligned} v &= 8.5867(2.0089^3) – 6.44(2.0089) + 4 \\ &= 60.67 ft/s \\ &= 60.7 ft/s \end{aligned}

Kinematics is a subfield of physics, developed in classical mechanics, that describes the motion of points, bodies (objects), and systems of bodies (groups of objects) without considering the forces that cause them to move.

Kinematics, as a field of study, is often referred to as the "geometry of motion" and is occasionally seen as a branch of mathematics. A kinematics problem begins by describing the geometry of the system and declaring the initial conditions of any known values of position, velocity and/or acceleration of points within the system.

Then, using arguments from geometry, the position, velocity and acceleration of any unknown parts of the system can be determined. The study of how forces act on bodies falls within kinetics, not kinematics. For further details, see analytical dynamics.

Learn more about kinematics here : brainly.com/question/24486060

#SPJ4

5 0
1 year ago
A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.
kirill115 [55]

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight=\sqrt{\frac{2H}{g}}

substituting value

  • =\sqrt{\frac{2*25}{9.8}}
  • =2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u*\sqrt{\frac{2H}{g}}

  • =8.25*2.26
  • =18.63m

<h3><u>The ball strikes 18.63m far from building</u>. </h3>
7 0
2 years ago
Radio waves transmitted through space at 3.00 ✕ 108 m/s by the Voyager spacecraft have a wavelength of 0.137 m. What is their fr
fredd [130]

Answer:

Frequency, f=2.18\times 10^9\ Hz

Explanation:

We have,

Speed of radio waves is 3\times 10^8\ m/s

Wavelength of radio waves is \lambda=0.137\ m

It is required to find the frequency of the radio waves. The speed of a wave is given by :

v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz

So, the frequency of the radio wave is 2.18\times 10^9\ Hz.

8 0
2 years ago
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