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raketka [301]
4 years ago
7

WILL give brainliest!!!!

Physics
1 answer:
Lostsunrise [7]4 years ago
8 0
People who have been struck by lightning do not carry an electrical charge and can be touched is True.

heat lightning is lightning flashes in the sky that do not have the accompanying sound of thunder so false just like the others 
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What is the angle o in the triangle below? (Hint: Use an inverse trigonometric<br> function.)
Shalnov [3]

Answer:

D. 66.4

Explanation:

So this problem uses SOHCAHTOA or the three trig functions.

Specifically this uses cosine, because it has an adjacent and a hypotenuse.

First you would determine what to do on the calculator, and since the problem is saying so, use the inverse cosine button. This will give you a angle measure from the decimal.

On a calculator, type in cos^-1(6/15). I used 6/15 because cosine is adjacent over hypotenuse. This will give you 66.4, which is D on the answers.

5 0
4 years ago
A metal ring 4.00 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpend
liq [111]

Answer:

a) 0.21N/C

b) counterclockwise

Explanation:

a) to find the magnitude of the electric field you can use the following formula:

\int Eds=-\frac{\Delta \Phi_B}{\Delta t}=-\frac{\Delta AB}{\Delta t}

A: area of the ring = pi*r^2

E: electric field

Ф_B: magnetic flux

In the line integral you can assume E as constant. Furthermore, you calculate the change in the magnetic flux by taking into account that the time interval is 1.12/0.21=5.33s. By replacing in the formula you obtain:

\frac{\Delta \Phi_B}{\Delta t}=\frac{A(B_f-B_i)}{5.33s}=\frac{\pi(0.04m)^2(1.12T)}{5.33}=1.056*10^{-3}W/s

E\int ds=E(2\pi r)=1.056*10^{-3}W/s\\\\E=\frac{1.056*10^{-3}W/s}{\pi(0.04m)^2}=0.21\frac{N}{C}

the magnitude if the induced electric field is 0.21N/C

b) By the Lenz's law you can conclude that the current has a direction in a counterclockwise

6 0
3 years ago
The stoplight had just changed and a 2200 kg cadillac had entered the intersection, heading north at 2.8 m/s , when it was struc
tekilochka [14]
<span>3. The attempt at a solution So basically what I did was divided into components. x: (3)(2000) = (3000)*v_x y: (v_vw)*(10000) = (3000)*v_y v_x, v_y is the velocity (after collision) in the x and y direction, respectively, of both cars stuck together (since it is an inelastic collision). v_vw is the initial velocity of the Volkswagen. Now what I did was that the angle is 35 degrees north of east. So basically made a triangle and figured that tan(35) = (v_y)/(v_x). This means (v_x)*(tan35) = v_y. Then, I simplified the component equations to get: x: 2 = v_x y: v_vw = 3*v_y Then plugging in for v_y, I got: v_vw = 3(2)(tan35) = 4.2 m/s as the velocity of the volkswagen. However, the answer key says 8.6 m/s. Could someone please help me out? Thanks Phys.org - latest science and technology news stories on Phys.org • Game over? Computer beats human champ in ancient Chinese game • Simplifying solar cells with a new mix of materials • Imaged 'jets' reveal cerium's post-shock inner strength Oct 24, 2012 #2 ehild Homework Helper Gold Member What directions you call x and y? Reference https://www.physicsforums.com/threads/2d-momentum-problem.646613/</span>
7 0
3 years ago
A thin, horizontal square copper plate has a length of 6.35cm and is charged with 2.56 × 108 electrons. If the electrons are uni
Strike441 [17]
This is either a trick question or a very hard one. In the first case: An electric field cannot occur inside a conductor (or by using the superposition principle you find out that at this point your electric fields cancel each other out)... Or you must use the laplace equation and proper boundary conditions to solve for the electrostatic potential 

<span>Edit: Considering the center of mass of the plate is on the plate</span>
5 0
3 years ago
The first moment of the total cross-sectional area taken about the neutral axis must be?
bekas [8.4K]

The first moment of the total cross sectional area taken about the neutral axis must be zero.

As with non-composite beams, the neutral axis (NA) is the location where the bending stress is zero. The location of the NA depends on the relative stiffness and size of each of the material sections.

Generally, the NA location is determined relative to the bottom surface of the beam. However, this is not mandatory, and the location can be relative to any location. If the bottom is used, then the NA axis is a distance "h"

The distance h can be determined by recalling that the stresses through the cross section must be in equilibrium.

learn more about neutral axis from here: brainly.com/question/28167877

#SPJ4

7 0
2 years ago
Read 2 more answers
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