Ask question

Ask question

All categories

3 years ago

9

When a ray strikes concave or convex mirrors obliquely at its pole. It's forming ______ straight angle right angle obtuse angle

acute angle

1 answer:

Andreas93 [3]3 years ago

4 0

Answer:

Straight Angle.

Explanation:

The two types of mirrors, i.e., concave mirror and convex mirror, acts in the same way when a ray is directed towards its pole, and the ray bounces back with the same intensity, and we get the exact same angle.

And, the initial angle and the reflected angle are exactly the same.

Hence, when the ray strikes the pole, in straight angle, then it forms a straight angle.

Hence, from the given option,

The correct answer is straight angle.

You might be interested in

The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

Law Incorporation [45]

The gravitational force between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is $r=6940 km=6.94 \cdot 10^6 m$ , the gravitational force is $F=9.21 \cdot 10^4 N$ and the mass of the Earth is $m_1=5.98 \cdot 10^{24} kg$ , therefore we can rearrange the previous equation to find m2, the mass of the telescope:
$m_2 = \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg$

6 0

3 years ago

Read 2 more answers

what was the acceleration of the cart with Low fan speed cm/s squared? what was the acceleration of the cart with medium fan spe

ArbitrLikvidat [17]

Explanation:

The attached figure shows data for the cart speed, distance and time.

For low fan speed,

Distance, d = 500 cm

Time, t = 7.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{7.4}\\\\v=67.56\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{67.56}{7.4}\\\\a=9.12\ cm/s^2$

For medium fan speed,

Distance, d = 500 cm

Time, t = 6.4 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{6.4}\\\\v=78.12\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{78.12}{6.4}\\\\a=12.2\ cm/s^2$

For high fan speed,

Distance, d = 500 cm

Time, t = 5.6 s

Average velocity,

$v=\dfrac{d}{t}\\\\v=\dfrac{500}{5.6}\\\\v=89.28\ cm/s$

Acceleration,

$a=\dfrac{v}{t}\\\\a=\dfrac{89.28}{5.6}\\\\a=15.94\ cm/s^2$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

Using a 683 nm wavelength laser, you form the diffraction pattern of a 1.1 mm wide slit on a screen. You measure on the screen t

n200080 [17]

Answer:

10.2 m

Explanation:

The position of the dark fringes (destructive interference) formed on a distant screen in the interference pattern produced by diffraction from a single slit are given by the formula:

$y=\frac{\lambda (m+\frac{1}{2})D}{d}$

where

y is the position of the m-th minimum

m is the order of the minimum

D is the distance of the screen from the slit

d is the width of the slit

$\lambda$ is the wavelength of the light used

In this problem we have:

$\lambda=683 nm = 683\cdot 10^{-9} m$ is the wavelength of the light

$d=1.1 mm = 0.0011 m$ is the width of the slit

m = 13 is the order of the minimum

$y=8.57 cm = 0.0857 m$ is the distance of the 13th dark fringe from the central maximum

Solving for D, we find the distance of the screen from the slit:

$D=\frac{yd}{\lambda(m+\frac{1}{2})}=\frac{(0.0857)(0.0011)}{(683\cdot 10^{-9})(13+\frac{1}{2})}=10.2 m$

6 0

3 years ago

14. The design for a rotating spacecraft below consists of two rings. The outer ring with a radius of 30 m holds the living quar

Zolol [24]

Answer:

$T= 11.0003$ s

Explanation:

From the question we are told that

The outer ring with a radius of 30 m

inner Gravity Approximately 9.80 m/s'

Outer Gravity Approximately 5.35 m/s.

Generally the equation for centripetal force is given mathematically as

Centripetal acceleration enables Rotation therefore?

$\omega ^2 r =Angular\ acc$

Considering the outer ring,

$\omega ^2 r = 9.8$

$\omega ^2= \frac{9.8}{30}$

$\omega = \sqrt{\frac{9.8}{30}}$

$\omega= 0.571 rad/s$

Therefore solving for Period T

Generally the equation for solving Period T is mathematically given as

$T= \frac{2\pi}{\omega}$

$T= \frac{2\pi}{0.571 rad/s}$

$T= 11.0003$ s

5 0

3 years ago

For this car, predict the shape of a graph that shows distance (x) versus time (t). Note that time is the independent variable a

Mrrafil [7]

Hope this helps!

ANSWER:

5 0

4 years ago