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xxTIMURxx [149]
2 years ago
13

3) Principles of rectilinear proportion of light 9​

Physics
1 answer:
8_murik_8 [283]2 years ago
7 0

Answer:Principle of rectilinear propagation of light

Explanation:Principle of rectilinear propagation of light

Rectilinear propagation of light refers to the propensity of light to travel along a straight line without any interference in its trajectory. ... It is because light travels along a straight line and leaves only the areas where the object interferes.

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A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of
balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

∴ f=\mu mg=0.490\times 90\times 9.8=432.18\ N

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

W=-fd=-432.18d

Now, change in kinetic energy is given as:

\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J

Therefore, from work-energy theorem,

W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m

Hence, the skater covers a distance of 15 m before stopping.

7 0
3 years ago
Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb
GuDViN [60]

Answer:

Q = 12540  J

Explanation:

It is given that,

Mass of water, m = 50 mL = 50 g

It is heated from 0 degrees Celsius to 60 degrees Celsius.

We need to find the energy required to heat the water. The formula use to find it as follows :

Q=mc\Delta T

Where c is the specific heat of water, c = 4.18 J/g°C

Put all the values,

Q=50\times 4.18\times (60-0)\\Q=12540\ J

So, 12540 J of energy is used to heat the water.

7 0
2 years ago
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