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laiz [17]
3 years ago
10

Why can you see your own reflected image in a mirror but not on a dry, painted wall?

Physics
1 answer:
kaheart [24]3 years ago
7 0
A wall uses diffuse reflection while a mirror uses specular reflection. For example, when parallel light rays enter a mirror, they remain parallel when exiting the mirror, allowing you to see a reflection of the light rays. On the contrary, when incident light rays enter a wall which is painted, the rays scatter, not allowing you to see anything but a painted wall. 
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Small blocks A and B are held at rest on a smooth plane inclined at 30° to the horizontal. Each is held in equilibrium by a forc
Olin [163]
<h3><u>Answer</u> :</h3>

First of all, See the attachment for better understanding.

<u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u><u>_</u>

❂ <u>Weight of block A</u> :

➝ mg sin30° = 18

➝ W (1/2) = 18

➝ W = 18×2

➝ <u>W = 36N </u>

❂ <u>Weight of block B</u> :

➝ N sin30° = 18

➝ (Mg cos30°) sin30° = 18

➝ W' (√3/2)(1/2) = 18

➝ W' (√3/4) = 18

➝ W' = 72/√3

➝ <u>W' = 41.61N</u>

5 0
3 years ago
Explain how to measure volume using a graduated cylinder
Reil [10]
Put object into cylinder and notice the volume increased. Difference of volume is the object's volume
6 0
3 years ago
+
nika2105 [10]

Answer:

The representation of a chemical reaction in the form of substances is known as a chemical equation. The equation in which the number of atoms of all the molecules is equal on both sides of the equation is known as a balanced chemical equation.

The Law of conservation of mass governs the balancing of a chemical equation.

According to this law, mass can neither be created nor be destroyed in a chemical reaction and obeying this law total mass of the elements or molecules present on the reactant side should be equal to the total mass of elements or molecules present on the produc

4 0
2 years ago
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Answer:

Im pretty sure its b

4 0
3 years ago
You throw a bowling ball of mass M and radius 1m has an initial speed v0 down the lane. It has an initial angular velocity of 3v
zhuklara [117]

Answer:

Explanation:

Given

Initial angular velocity \omega_0=\frac{3v_0}{R}

Initial speed u=v_0

Coefficient of kinetic friction \mu _k=0.08

As No external torque is applied therefore angular momentum is conserved

L_i=L_f

I_{cm}v_i+Mv_{cm}_iR=I_{cm}\omega _f+Mv_{cm}_f\cdot R

For rolling without slipping v_f=\omega _fR

\frac{2}{5}MR^2\cdot (\frac{3v_0}{R})+Mv_0\cdot R=\frac{2}{5}MR^2\cdot \frac{v_f}{R}+Mv_f\cdot R

\frac{6}{5}Mv_0R+Mv_0R=\frac{2}{5}Mv_fR+Mv_f\cdot R

v_f=\frac{11}{7}v_0

acceleration provided by surface

a=\mu _kg

a=0.08\times 9.8

a=0.784\ m/s^2

Speed of ball after t=2\ s

using v=u+at

v=final velocity

u=initial velocity

\frac{11}{7}v_0=v_0+0.784\times 2

\frac{11}{7}v_0-v_0=0.784\times 2

v_0=2.744\ m/s

thus v_f=\frac{11}{7}\times 2.74

v_f=4.312\ m/s

5 0
3 years ago
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