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3 years ago

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For using the law of reflection and Snell's law for refraction, the angles are measured Group of answer choices between the surf

ace and the light ray between the normal to the surface and the light ray either way of measuring the angle will work for both reflection and refraction

1 answer:

Scrat [10]3 years ago

8 0

Answer:

correct answer is the c

Explanation:

The laws of optical geometry are the law of reflection and refraction, in them the relationship between the incident angles and reflected or transmitted by a given surface is established, in these laws it is not specified how the angles should be measured by which they could measure as follows

.- between the beam and the surface

.- between lightning and normal, the most used

Let's check the answers

a) True.

b) True

c) True

We see that the three answers are true, but the answer c involves the other two, so the correct answer is the c

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If a bust starts to move and its velocity becomes 90 km after 8 seconds . calculate its acceleration answer it quick please

kherson [118]

Answer:

a = 3.125 [m/s^2]

Explanation:

In order to solve this problem, we must use the following equation of kinematics. But first, we have to convert the speed of 90 [km/h] to meters per second.

$90\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s} \\= 25 \frac{m}{s}$

$v_{f} =v_{i} + (a*t)$

where:

Vf = final velocity = 25 [m/s]

Vi = initial velocity = 0

a = acceleration [m/s^2]

t = time = 8 [s]

The initial speed is zero as the bus starts to koverse from rest. The positive sign of the equation means that the bus increases its speed.

25 = 0 + a*8

a = 3.125 [m/s^2]

3 0

3 years ago

A bullet is fired into the air at an angle of 45°. How far does it travel before it is 1,000 feet above the ground? (Assume that

Readme [11.4K]

Answer:

It travels 1414 feets.

Explanation:

Let's take the length the bullet travels <em>l </em>as the hypotenuse of a right triangle and the height it reaches one of its sides. Since we got the angle α at which it was fired and the height <em>h</em> it reached, we can calculate <em>l</em> using the <em>sin(α)</em> function:

$sin(\alpha )=\frac{opposite side}{hypotenuse}\\sin(\alpha)=\frac{h}{l}\\l=\frac{h}{sin(\alpha)}$

Replacing:

$l=\frac{1000ft}{sin(\frac{\pi}{4})}$

Solving and roundin to the nearest foot:

$l=1414 ft$

3 0

3 years ago

A large crane consists of a 20 m, 3,000 kg arm that extends horizontally on top of a vertical tower. The arm extends 15 m toward

Anni [7]

Answer:

$m=18000kg$

Explanation:

From the question we are told that:

Crane Length $l=20m$

Crane Mass $m_a=3000kg$

Arm extension at lifting end $l_l=15m$

Arm extension at counter weight end $l_c=5m$

Load $M_l=5000kg$

Generally the equation for Torque Balance is mathematically given by

$T_1 *l_c-(m_a*g) *l_c-(T_2)*l_l=0$

$mg*5 *-(3000*9.8) *5-(5000*9.8)*15=0$

$m=18000kg$

7 0

3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0

3 years ago

A teacher walks 5m north of his desk, then we walks 6m south.

sergey [27]

11m if you add 6+5 you get 11 but of course you need the “m” in the mix so 11m but correct me if I’m wrong.

8 0

3 years ago