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Andreyy89
2 years ago
10

Electric fields are vector quantities whose magnitudes are measured in units of volts/meter (V/m). Find the resultant electric f

ield when there are two fields, E1and E2, where E1 is directed vertically upward and has magnitude 100 V/m and E2 is directed 45 degrees to the left of E1 and has magnitude 150 V/m. Use a graph to show vector drawing!

Physics
1 answer:
musickatia [10]2 years ago
8 0

Answer:

Er = 231.76 V/m, 27.23° to the left of E1

Explanation:

To find the resultant electric field, you can use the component method. Where you add the respective x-component and y-component of each vector:

E1:

E_1_x = 0V/m\\E_1_y=100V/m

E2:

Keep in mind that the x component of electric field E2 is directed to the left.

E_2_x= 150V/m*-sin(45) = 106.07 V/m\\E_2_y=150V/m*cos(45) = 106.07V/m

∑x: E_1_x+E_2_x = 0V/m - 106.07V/m = -106.07V/m

∑y: E_1_y + E_2_y = 100V/m + 106.07V/m = 206.07V/m

The magnitud of the resulting electric field can be found using pythagorean theorem. For the direction, we will use trigonometry.

||E_r||= \sqrt{(-106.07V/m)^2+(206.07V/m)^2} = 231.76 V/m\\\\\alpha = arctan(\frac{206.7 V/m}{-106.07 V/m}) = 117.24degrees

or 27.23° to the left of E1.

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Grace [21]

Answer:

4.3 * 10 N

Explanation:

To calculate torque, we multiply the distance from the pivot by the perpendicular (the part of the force that acts at right angles to the displacement vector) component of the force to the displacement vector from the pivot.

torque  = distance from pivot *  perpendicular force

170 Nm= 0.4 m * F

F = 425 N = 4.3 * 10 N rounded off to two significant figures

4 0
2 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

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Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
Find the shear stress and the thickness of the boundary layer (a) at the center and (b) at the trailing edge of a smooth flat pl
melomori [17]

Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

The boundary layer thickness is equal to:

\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

The sear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

7 0
3 years ago
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Nataly_w [17]

I think its B B)Warm water rises and cold water moves in to replace it.

7 0
3 years ago
Read 2 more answers
1) A record is spinning at the rate of 25 rpm. If a ladybug is sitting 10 cm from the
Sladkaya [172]

<h2>distance = 523 cm</h2>

Explanation:

( a ) The rotational speed of the ladybug = 25 r.p.m = 25/60 r.p.s

= 5/12 rev/sec

( b ) The definition of frequency is the number of rotations per second .

Here the number of rotations per second is 5/12 . Thus frequency = 5/12 Hz

( c ) The tangential speed is v = angular velocity x radius of rotation

The angular velocity ω = 2π x n , where n is the number of rotations per second

Thus angular velocity = 2π x 5/12   = 5π/6 rad/sec

The linear velocity = angular velocity x distance from center of record

Thus tangential speed = 5π/6 x 10 = 25π/3 cm/sec

Angular displacement in 20 sec = ω x t = 5π/6 x 20 = 50π/3 rad

Linear displacement = angular displacement x distance from center of record

= 50π/3 x 10 = 500π/3 = 523 cm

8 0
3 years ago
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