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ANEK [815]
3 years ago
11

I NEED THIS ANSWER ASAP PLEASE!!! What happens when a light ray passes through the focal point, and then reflects off of a conve

x mirror? A.The reflected ray will run parallel to the center of curvature. B. The reflected ray will run parallel to the principal axls. C. The reflected ray will pass through the center of curvature. D. The reflected ray will pass through a point between the center of curvature and the focal point. E. The reflected ray will pass through a point between the focal point and the vertex
Physics
1 answer:
Alexandra [31]3 years ago
8 0

What happens when a light ray passes through the focal point, and then reflects off of a convex mirror?

E. The reflected ray will pass through a point between the focal point and the vertex

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Can someone help please and thank you:)
Ray Of Light [21]

Answer:

The answer is C.

Explanation:

4 0
2 years ago
A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (
SIZIF [17.4K]

The solution to the questions are given as

  • t=40.39 \mathrm{sec}
  • \varepsilon &=(0.12v)e^{0.057t}
  • the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}

\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$

\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}

\varepsilon &=(0.12v)e^{0.057t}

(b) Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$

\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

brainly.com/question/13076734

#SPJ1

4 0
2 years ago
What is the force on a 1000kg elevator that is falling freely at 9.8 m/sec^2
Advocard [28]
From Newton's second law, we know F = ma, where a is the acceleration and m is the mass in kg.

F = 1000kg * 9.8m/s = 9800N

F = 9800 N

Hope this helps!
3 0
2 years ago
Part 1 :
tatuchka [14]

1) 9.57 N

We have two forces applied on the apple:

- The force of gravity, in the downward direction:

W = 9.42 N

- The force exerted by the wind, in the horizontal direction (to the right):

Fw = 1.68 N

The two forces are perpendicular to each other, so we can find the magnitude of the net force by using Pythagorean's theorem.

Therefore, we have:

F=\sqrt{W^2+F_w^2}=\sqrt{(9.42)^2+(1.68)^2}=9.57 N

2) 10^{\circ}

The direction of the net external force, measured from the downward vertical, can be measured using the following formula:

\theta = tan^{-1}(\frac{F_x}{F_y})

where

F_x is the force in the horizontal direction

F_y is the force in the vertical direction

In this problem,

F_x = F_w = 1.68 N

F_y = W = 9.42 N

and so we find:

\theta = tan^{-1}(\frac{1.68}{9.42})=10^{\circ}

4 0
2 years ago
A caterpillar tries to climb straight up a wall two meters high, but for every 2 cm up it climbs, it slides down 1 cm. Eventuall
MaRussiya [10]

Answer:

this question does not make sense could you please make it clearer

3 0
2 years ago
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