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Gnoma [55]
3 years ago
7

Which of the following statements are true for electric field lines? Check all that apply. Check all that apply. Electric field

lines point away from positive charges and toward negative charges. Electric field lines are continuous; they do not have a beginning or an ending. Electric field lines can never intersect. Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. At every point in space, the electric field vector at that point is tangent to the electric field line through that point.
Physics
1 answer:
Scilla [17]3 years ago
5 0

Answer:

Electric field lines point away from positive charges and toward negative charges. <u>True</u>

Electric field lines are continuous; they do not have a beginning or an ending.<u> False</u>

Electric field lines can never intersect. <u>True</u>

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

Explanation:

Electric field lines point away from positive charges and toward negative charges. Always the field lines go to negative charges and leave from positive charges.

Electric field lines are continuous; they do not have a beginning or an ending.<u> False  </u>

Because the field lines starts at positive charges and ends in negative charges.

Electric field lines can never intersect. <u>True</u>

It can not intercept the field lines because in that point the the field would have two directions. Besides, in that point the real value of the field should be found adding both field lines.

Electric field lines are close together in regions of space where the magnitude the electric field is weak and are father apart where it is strong. <u>False</u>

This fact is opposite to that so the regions of space where the magnitude the electric field is weak the lines are father apart and where the field is strong  the lines are close together.

At every point in space, the electric field vector at that point is tangent to the electric field line through that point.<u> True</u>

This statement correspond to the definition of the field line.

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A rock is thrown at a window that is located 18.0 m above the ground. The rock is thrown at an angle of 40.0° above horizontal.
Korvikt [17]

Answer:

B) 27.3 m

Explanation:

The rock describes a parabolic path.

The parabolic movement results from the composition of a uniform rectilinear motion (horizontal ) and a uniformly accelerated rectilinear motion of upward or downward motion (vertical ).

The equation of uniform rectilinear motion (horizontal ) for the x axis is :

x =  vx*t   Equation (1)

Where:  

x: horizontal position in meters (m)

t : time (s)

vx: horizontal velocity  in m/s  

The equations of uniformly accelerated rectilinear motion of upward (vertical ) for the y axis  are:

(vfy)² = (v₀y)² - 2g(y- y₀)    Equation (2)

vfy = v₀y -gt    Equation (3)

Where:  

y: vertical position in meters (m)  

y₀ : initial vertical position in meters (m)  

t : time in seconds (s)

v₀y: initial  vertical velocity  in m/s  

vfy: final  vertical velocity  in m/s  

g: acceleration due to gravity in m/s²

Data

v₀ = 30 m/s , at an angle  α=40.0° above the horizontal

v₀x = vx = 30*cos40° = 22.98 m/s

v₀y = 30*sin40° = 19.28 m/s

y₀ = 2m

y =  18.0 m

g = 9.8 m/s²

Calculation of the time (t) it takes for the rock to reach at  18 m above the ground

We replace data in the equation (2)

(vfy)² = (v₀y)² - 2g(y- y₀)    

(vfy)² = (19.28)² - 2(9.8)(18- 2)

(vfy)² = 371.86 - 313.6

(vfy)² = 58.26

v_{f} = \sqrt{58.26}

vfy = 7.63 m/s

We replace vfy = 7.63 m/s in the equation (2)

vfy = v₀y - gt

7.63 = 19.28 - (9.8)(t)

(9.8)(t) = 11.65

t = 11.65 / (9.8)

t = 1.19 s

Horizontal distance from where the rock was thrown to the window

We replace t = 1.19 s , in the equation (1)

x =  vx*t  

x = (22.98)* ( 1.19 )

x = 27.3 m

3 0
3 years ago
Does uranium use steam?
Neko [114]

Answer:

Yes.

Explanation:

Reactors use uranium for nuclear fuel. The uranium is processed into small ceramic pellets and stacked together onto sealed metal tubes called fuel rods. The heat created by fission turns the water into steam.

4 0
2 years ago
The reflective surface of a CD consists of spirals of equally spaced grooves. If you shine a laser pointer on a CD, each groove
Ipatiy [6.2K]

Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, d = \frac{m \lambda}{sin(A_{m}) }..........(1)

The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, \lambda = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

A_{m} = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}

sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}

Putting all appropriate values into equation (1)

d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m

3 0
3 years ago
A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure
kozerog [31]

Answer:

43.7 °C

Explanation:

\alpha_b = Coefficient of linear expansion of brass = 18\times 10^{-6}\ ^{\circ}C

\alpha_s = Coefficient of linear expansion of steel = 11\times 10^{-6}\ ^{\circ}C

L_{0b} = Initial length of brass = 31 cm

L_{0s} = Initial length of steel = 11 m

\Delta L = Total change in length = 3 mm

Total change in length would be

\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C

\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C

The final temperature is 43.7 °C

6 0
3 years ago
3. a) If the distance between two objects increases by a factor of 5 (5),
irina [24]

Answer:

Answer for A

Explanation:

F1=GmM/r1^2

If r2 becomes r2=5r

F2=GmM/(25r^2)

Multiply with 25 gives to maintain the same force

I.e.,25F2=F1

F2=G(25m)M/25r^2=F1

By the factor 25 would change to increase to same.

3 0
2 years ago
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