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xz_007 [3.2K]
3 years ago
13

HELP HELP HELP ME!!!

Physics
1 answer:
zzz [600]3 years ago
7 0
I believe the answer is C. Hope this helps!!

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Would a rougher or smoother object absorb thermal energy faster? Why? Give an example.
Nuetrik [128]
Smoother because it will increase energy and when the energy increases it’ll create heat also . Example: A car racing on a smooth road it’ll go faster than a Car speeding on a bumpy and rough road , Hope that helps .
5 0
3 years ago
Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

5 0
3 years ago
Why is a protective apron or lab coat important to use when working with acids?
Oksanka [162]

Answer: Option (C) is the correct answer.

Explanation:

Whether an acid is weak or strong it will cause harm if it comes in direct contact with the skin, eyes or mouth etc.

Thus, in order to prevent oneself, it is advised to wear a lab coat while working in laboratory as acids break down fabrics and can cause burns if the acids are strong.

Hence, a lab coat acts as a protective layer on the clothes so that even if acid falls on it then it might not reach the skin immediately and on time prevention can be taken.

6 0
3 years ago
Read 2 more answers
Einstein and Lorentz, being avid tennis players, play a fast-paced game on a court where they stand 20.0 m from each other. Bein
igor_vitrenko [27]

Answer:

A) 185.6 J

B) 9.396 x 10^14 J

C) 4x10^7 m/s

D) 20 m

E) 9.09x10^-8 sec

F) 9.09x10^-8 sec

Explanation:

Detailed explanation and calculation is shown in the image below

3 0
3 years ago
10. Calculate A ramp is
inna [77]

Answer: 3.33

Explanation:

MA = De/Dr

MA = 5.0/1.5

MA= 3.33

5 0
4 years ago
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