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3 years ago

If the angle of reflection is 82 and the light ray strikes a flat mirror, what will the angle

Physics

2 answers:

omeli [17]3 years ago

6 0

Answer:

The angle of incidence will be equal to the reflection.

Explanation:

The law of reflection states that when a ray of light reflects off a surface, the angle of incidence is equal to the angle of reflection.

So 82 would be the answer.

il63 [147K]3 years ago

4 0

Answer:

Explanation:

According to law of reflection, the angle of incidence = angle of reflection

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A 1.50 x 109 kg railroad car moving at 7.00 m/s to the north collides with

barxatty [35]

GIVE ME YOUR MONEY

Explanation:

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3 years ago

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L) is defined as:

$T=2\,\pi\,\sqrt{ \frac{L}{g} }$

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

$T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2$

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0

3 years ago

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on

leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

$q_1=50\ nC at y=6\ m$

$q_2=-80\ nC at x=-4\ m$

$q_3=70\ nC at y=-6\ m$

Electric Potential is given by

$V=\frac{kQ}{r}$

Distance of $q_1$ from $x=8\ m$

$d_1=\sqrt{6^2+8^2}$

$d_1=\sqrt{36+64}$

$d_1=10\ m$

similarly $d_2=8-(-4)$

$d_2=12\ m$

$d_3=\sqrt{(-6)^2+8^2}$

$d_3=\sqrt{36+64}$

$d_3=10\ m$

Potential at $x=8\ m$ is

$V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}$

$V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]$

$V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}$

$V_{net}=9\times 5.33$

$V_{net}=47.97\approx 48\ V$

5 0

3 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f

Hitman42 [59]

(a) The angular acceleration of the wheel is given by
$\alpha = \frac{\omega_f - \omega_i }{t}$
where $\omega_i$ and $\omega_f$ are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
$\alpha = \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2$

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
$\theta (t) = \omega_i t + \frac{1}{2} \alpha t^2$
where $\omega_i = 0$ is the initial angular speed. So, the angle covered after a time t=3.07 s is
$\theta= \frac{1}{2} \alpha t^2 = \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad$

6 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m