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ladessa [460]
3 years ago
9

What happens to light when it strikes the inside surface of a smooth, curved mirror? It bounces in random directions. It bounces

back toward a single spot. It passes through the mirror and moves in random directions. It passes through the mirror and moves in a straight line.
Physics
2 answers:
kotykmax [81]3 years ago
5 0
This is a concave mirror you're talking about, so all of the points are going to converge to a single focal point. Therefore the answer would be that it bounces back toward a single spot
tester [92]3 years ago
4 0

Answer:

It bounces back towards a single spot.

Explanation:

When the light ray strikes to the inner surface of smooth, curved mirror (concave mirror), it bounces back at a single point and the point is called the focal point of the concave mirror.

When a ray of light parallel to principal axis strikes to the concave mirror, then after reflection it goes through the focus, i.e., at a single spot.

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D protons

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A circuit has a voltage drop of 24.0 V across a 30.0 resistor that carries a
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(A) 19.2 W

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The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second
just olya [345]

Answer:

g=274\ m/s^2

Explanation:

Mass of the Sun, M=1.99\times 10^{30}\ kg

The radius of the Sun, r=6.96\times 10^8\ m

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

g=\dfrac{GM}{r^2}\\\\g=\dfrac{6.67\times 10^{-11}\times 1.99\times 10^{30}}{(6.96\times 10^8)^2}\\\\g=274\ m/s^2

So, the value of acceleration due to gravity on the Sun is 274\ m/s^2.

8 0
2 years ago
A hawk is flying horizontally at 18.0 m/s in a straight line, 230 m above the ground. A mouse it has been carrying struggles fre
Lisa [10]

Answer:

a) vd = 47.88 m/s

b) θ = 80.9°

c) t = 6.8 s

Explanation:

In the situation of the problem, you can assume that the trajectory of the hawk and the trajectory of the mouse form a rectangle triangle.

One side of the triangle is the horizontal trajectory of the hawk after 2.00s of flight, the other side of the triangle is the distance traveled by the mouse when it is falling down. And the hypotenuse is the trajectory of the hawk when it is trying to recover the mouse.

(a) In order to calculate the diving speed of the hawk, you first calculate the hypotenuse of the triangle.

One side of the triangle is c1 = (18.0m/s)(2.0s) = 36m

The other side of the triangle is c2 = 230m - 3m = 227 m

Then, the hypotenuse is:

h=\sqrt{(36m)^2+(227m)^2}=229.83m    (1)

Next, it is necessary to calculate the falling down time of the mouse, this can be done by using the following formula:

y=y_o+v_ot+\frac{1}{2}gt^2    (2)

yo: initial height = 230m

vo: initial vertical speed of the mouse = 0m/s

g: gravitational acceleration = -9.8m/s^2

y: final height of the mouse = 3 m

You replace the values of the parameters in (2) and solve for t:

3=230-4.9t^2\\\\t=\sqrt{\frac{227}{4.9}}=6.8s

The hawk traveled during 2.00 second in the horizontal trajectory, hence, the hawk needed 6.8s - 2.0s = 4.8 s to travel the distance equivalent to the hypotenuse to catch the mouse.

You use the value of h and 4.8s to find the diving speed of the hawk:

v_d=\frac{229.83m}{4.8s}=47.88\frac{m}{s}

The diving speed of the Hawk is 47.88m/s

(b) The angle is given by:

\theta=cos^{-1}(\frac{c_1}{h})=cos^{-1}(\frac{36m}{229.83m})=80.9 \°

Then angle between the horizontal and the trajectory of the Hawk when it is descending is 80.9°

(c) The mouse is falling down during 6.8 s

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2e min :)) pls park braliest
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