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CaHeK987 [17]
3 years ago
6

At what point will the electric field of a charged object be strongest?

Physics
2 answers:
Y_Kistochka [10]3 years ago
7 0

Answer:

C. Where the electric field lines are closest together

Assuming this question is from Plato.

Explanation:

maks197457 [2]3 years ago
6 0
The electric field of a charged object will be strongest when all the field lines are close to one another.

beacsue, the strength of electric field depends on the number of lines of electric field. More lines at a place, higher the electric field strength. Thus, any charge in that area will face more electric field.
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the inertial mass of an object is measured by exerting a force on the object and measuring the object's BLANK using an inertial
JulijaS [17]

Answer:

acceleration

sorry i answered late

3 0
3 years ago
A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an
ella [17]

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

v = w/k

99.6/1.25 = 79.68

s = vt

= 79.68 * 0.0683

= 5.02

Read more on waves here

brainly.com/question/25699025

#SPJ4

complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

7 0
1 year ago
Need help ASAP
kati45 [8]

Answer:

Explanation:

1.electric

2.radio wave

3.broadcast

Hope it helped you.

8 0
2 years ago
A 3 mm inside diameter tube is placed in a fluid with a surface tension of 600 mN/m and density of 3.7 g/cm3. The contact angle
Aleks04 [339]

Answer: The height of the fluid rise is 0.01m

Explanation:

Using the equation

h = (2TcosѲ )/rpg

h= height of the fluid rise

diameter of the tube =3mm

radius of the tube= 3/2 =1.5mm=0.0015

T= surface tension = 600mN/m=0.6N/m

Ѳ = contact angle = 60^oC

p= density =3.7g/cm3= 3700kg/m3

g= acceleration due to gravity =9.8m/s2

h = ( 2*0.6*0.5)/(0.0015*3700*9.8)

h = 0.6/54.39

h= 0.01m

Therefore,the height of the fluid rise is 0.01m

8 0
3 years ago
A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the
podryga [215]

Answer:

6.9\times 10^{28}m^{-3}

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=rr=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m

1mm=10^{-3} m

Current=I=8 A

Drift velocity=v_d=5.4\times 10^{-5} m/s

We have to find the density of free electrons in the metal

We know that

Density of electron=n=\frac{I}{v_deA}

Using the formula

Density of free electrons=\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}

By using Area of wire=\pi r^2

\pi=3.14\\e=1.6\times 10^{-19} C

Density of free electrons=6.9\times 10^{28}m^{-3}

3 0
3 years ago
Read 2 more answers
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