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Sveta_85 [38]
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
Physics
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.

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What is physical change ?​
Sidana [21]

Answer:

A physical change is a change to the physical—as opposed to chemical—properties of a substance. They are usually reversible. The physical properties of a substance include such characteristics as shape (volume and size), color, texture, flexibility, density, and mass.

Explanation:

PLS MARK AS THE BRAINLIEST

8 0
3 years ago
Read 2 more answers
A sled of mass m is given a kick on a frozen pond. The kick imparts to the sled an initial speed of 2.00 m/s . The coefficient o
kobusy [5.1K]

2.04 meters distance is traveled by the sled before stopping.

Mass of the sled = m

The initial speed of the sled = 2 m/s

Coefficient of kinetic friction between sled and ice = 0.100

Let the distance the sled moves before it stops be d.

Gravity = 9.8 m/ s²

Let the initial kinetic energy sled be

= K _{i}

K_{i} =  \frac{1}{2} mv ^{2}

The work done by the frictional force is,

Work \: done \:  by \: frictional \: force =W_{f}

W _{f} = μ_{k}mgd

Work done by frictional force= Initial kinetic energy of the sled

W_{f} = K_{i}

μ_{k}mgd= \frac{1}{2} mv ^{2}

So, the distance traveled by the sled before stopping is

d= \frac{1mv ^{2} }{2 \:μ_{k}mg}

d= \frac{1v ^{2} }{2 \:μ_{k}g}

d= \frac{2^{2} }{2  \times \:0.100 \times 9.8}

d= 2.04 \: m

Therefore, the distance traveled by the sled before stopping is 2.04 meters.

To know more about work done, refer to the below link:

brainly.com/question/13662169

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5 0
2 years ago
Please help me :) ty
slega [8]
1. Physical Property
2. Chemical Change
3. Chemical Change
4. Chemical Property
5. Physical Property
6. Physical Change
7. Chemical Change
8. Physical Change
9. Chemical Change
10. Physical Property
11. Physical Property
12. Chemical Change
13. Chemical Change
14. Chemical Property
15. Chemical Change

Explanation:
Physical Property: A property of matter that can be seen, felt, tasted, smelled, or heard.
Physical Change: The change of an object of matter's physical property/properties.
Chemical Property: An element/object of matter's internal property/properties
Chemical Change: An element/object of matter's internal property/properties being altered or changed

8 0
4 years ago
This is junior year english
pychu [463]
D. is the correct answer.
6 0
4 years ago
Read 2 more answers
An infinitely long line of charge has linear charge density 6.00×10−12 C/m . A proton (mass 1.67×10−27 kg,charge +1.60×10−19 C)
Bogdan [553]

Answer:

A) \,K.E=1.405\times 10^{-20}J

B)\,r_f=0.268\,m

Explanation:

Charge\,\,density=\lambda=6\times 10^{-12}C/n\\\\Mass\,\, of \,\,proton=m_p=1.67\times 10^{-27}kg\\\\charge\,\, of\,\, proton=q_p=1.609\times 10^{-19}C\\\\r=12\,cm=0.12\, m\\\\v=4.103\times 10^{3}m/s

A) Initial kinetic energy of proton

K.E=\frac{1}{2}m_pv^2\\\\K.E=1.405\times 10^{-20}J

B) How close does the proton get to the line of charge?

Potential energy and kinetic energy are related as:

K_i+U_i=K_f+U_f\\\\U_f-U_i=K_i-K_f\\\\q(V_f-V_i)=1.40\times 10^{-20}\\\\V_f-V_i=0.087--(1)\\

Change in voltage is

V_f-V_i=\frac{\lambda}{2\pi \epsilon_o}ln\frac{r_f}{r_i}\\\\ln|\frac{r_f}{r_i}|=(0.087)(\frac{2\pi \epsilon_o}{\lambda})\\\\ln|\frac{r_f}{r_i}|==0.8059\\\\\frac{r_f}{r_i}=2.24\\\\r_f=(2.24)(.12)\\\\r_f=0.268 m

5 0
3 years ago
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