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dimulka [17.4K]
2 years ago
8

I'm giving 40 points to whoever can answer my question!

Physics
1 answer:
QveST [7]2 years ago
5 0

How does the angle of incoming light compare to the angle of light reflecting off of the surface?

  • <em>The measurements of the angles are equal.</em>

<u>For</u><u> </u><u>a</u><u> </u><u>smooth</u><u> </u><u>surface</u><u>,</u><u> </u><u>The</u><u> </u><u>angle</u><u> </u><u>of</u><u> </u><u>reflection</u><u> </u><u>and</u><u> </u><u>the</u><u> </u><u>angle</u><u> </u><u>of</u><u> </u><u>incidence</u><u> </u><u>are</u><u> </u><u>equal</u><u>.</u>

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Why car speed is not considered as car velocity?​
grigory [225]

Answer:

Cause its scalar quantity

Explanation:

since speed does not take directions into consideration, it is considered to be a scalar quantity. On the other hand, the velocity of an object does not take into account direction, thus making it a vector quantity.

4 0
4 years ago
At what time was the person at a position of 0m?
Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line  equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

  = (0 x 1) + 1/2 (0 x 1 x 1)

  = (0) + 1/2 (0)

  = 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

 = 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1)  =  10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0
3 years ago
A _____ map provides information about how and when rocks formed in a particular area.A.contourB.geologicC.topographic
givi [52]

I believe the answer is B

6 0
3 years ago
It has been argued that power plants should make use of off-peak hours to generate mechanical energy and store it until it is ne
sdas [7]

Answer:

Explanation:

90 rpm = 90 / 60 rps

= 1.5 rps

= 1.5 x 2π rad /s

angular velocity of flywheel

ω = 3π rad /s

Let I be the moment of inertia of flywheel

kinetic energy = (1/2) I ω²

(1/2) I ω² = 10⁷ J

I = 2 x  10⁷ / ω²

=2 x  10⁷ / (3π)²

= 2.2538 x 10⁵ kg m²

Let radius of wheel be R

I = 1/2 M R² , M is mass of flywheel

= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .

1/2 πR⁴ x t x d = 2.2538 x 10⁵

R⁴ = 2 x 2.2538 x 10⁵ / πt d

= 4.5076 x 10⁵ / 3.14 x .1 x 7800

= 184

R= 3.683 m .

diameter = 7.366 m .

b ) centripetal accn required

= ω² R

= 9π² x 3.683

= 326.816 m /s²

3 0
3 years ago
WhAT is the change IN THE entropy of 2.0kg of h2o molecules when transform at a constant pressure of 1 atm from water at 100 deg
Fynjy0 [20]

Answer:

The entropy change is 45.2 kJ/K.

Explanation:

mass of water at 100 C = 2 kg

Latent heat of vaporization, L = 2260 kJ/kg

Heat is

H = m L

H = 2 x 2260 = 4520 kJ

Entropy is given by

S = H/T = 4520/100 = 45.2 kJ/K

3 0
3 years ago
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