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3 years ago

PLEASE HELP I need the answer to this question ASAP

Physics

2 answers:

cestrela7 [59]3 years ago

7 0

Answer:

answer is A

Explanation:

hope this helps!

Simora [160]3 years ago

7 0

Answer:

ray 2 is the refracted ray

Explanation:

ray 1 and 2 are balanced because if there is a mirror under the ray then the ray would be balance with the source of the ray. p.s only did this for brainliest

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What is the answer to this question

leva [86]

The answer is A 0 degrees is the freezing point of water

Plz mark brainliest

7 0

3 years ago

An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what facto

Lemur [1.5K]

Answer:

The factor of the diameter is 0.95.

Explanation:

Given that,

Power of old light bulb = 54.3 W

Power = 60 W

We know that,

The resistance is inversely proportional to the diameter.

$R\propto\dfrac{1}{D}$

The power is inversely proportional to the resistance.

$P\propto\dfrac{1}{R}$

$P\propto D^2$

We need to calculate the factor of the diameter of the filament reduced

Using relation of power and diameter

$\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}$

Put the value into the formula

$\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}$

$\dfrac{D_{i}}{D_{f}}=0.95$

$D_{i}=0.95 D_{f}$

Hence, The factor of the diameter is 0.95.

7 0

3 years ago

Read 2 more answers

A solid cylinder of mass M = 45 kg, radius R = 0.44 m and uniform density is pivoted on a frictionless axle coaxial with its sym

user100 [1]

Answer:

w_f = 1.0345 rad/s

Explanation:

Given:

- The mass of the solid cylinder M = 45 kg

- Radius of the cylinder R = 0.44 m

- The mass of the particle m = 3.6 kg

- The initial speed of cylinder w_i = 0 rad/s

- The initial speed of particle V_pi = 3.3 m/s

- Mass moment of inertia of cylinder I_c = 0.5*M*R^2

- Mass moment of inertia of a particle around an axis I_p = mR^2

Find:

- What is the magnitude of its angular velocity after the collision?

Solution:

- Consider the mass and the cylinder as a system. We will apply the conservation of angular momentum on the system.

L_i = L_f

- Initially, the particle is at edge at a distance R from center of cylinder axis with a velocity V_pi = 3.3 m/s contributing to the initial angular momentum of the system by:

L_(p,i) = m*V_pi*R

L_(p,i) = 3.6*3.3*0.44

L_(p,i) = 5.2272 kgm^2 /s

- While the cylinder was initially stationary w_i = 0:

L_(c,i) = I*w_i

L_(c,i) = 0.5*M*R^2*0

L_(c,i) = 0 kgm^2 /s

The initial momentum of the system is L_i:

L_i = L_(p,i) + L_(c,i)

L_i = 5.2272 + 0

L_i = 5.2272 kg-m^2/s

- After, the particle attaches itself to the cylinder, the mass and its distribution around the axis has been disturbed - requires an equivalent Inertia for the entire one body I_equivalent. The final angular momentum of the particle is as follows:

L_(p,f) = I_p*w_f

- Similarly, for the cylinder:

L_(c,f) = I_c*w_f

- Note, the final angular velocity w_f are same for both particle and cylinder. Every particle on a singular incompressible (rigid) body rotates at the same angular velocity around a fixed axis.

L_f = L_(p,f) + L_(c,f)

L_f = I_p*w_f + I_c*w_f

L_f = w_f*(I_p + I_c)

-Where, I_p + I_c is the new inertia for the entire body = I_equivalent that we discussed above. This could have been determined by the superposition principle as long as the axis of rotations are same for individual bodies or parallel axis theorem would have been applied for dissimilar axes.

L_i = L_f

5.2272 = w_f*(I_p + I_c)

w_f = 5.2272/ R^2*(m + 0.5M)

Plug in values:

w_f = 5.2272/ 0.44^2*(3.6 + 0.5*45)

w_f = 5.2272/ 5.05296

w_f = 1.0345 rad/s

5 0

3 years ago

2. We know that the Earth is moving around the sun. When we look at the sky we see

marin [14]

Answer:

Answer: The Sun, the Moon, the planets, and the stars all rise in the east and set in the west. And that's because Earth spins -- toward the east. ... Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

6 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