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harkovskaia [24]
3 years ago
13

Distinguish between speed and velocity?atleast 4 point should write​

Physics
1 answer:
Sergio [31]3 years ago
4 0

Answer:

mark me as brainliest ❤️

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The rhinestones in costume jewelry are glass with index of refraction 1.50. To make them more reflective, they are often coated
Sedaia [141]

Answer:

60

Explanation:

According to the given question, the computation of minimum coating thickness is shown below:-

The condition for constructive interference is

2t_{min} = (m + \frac{1}{2} )\times \frac{\lambda}{^nmateral}

= (0 + \frac{1}{2} )\times \frac{\lambda}{^nmateral}

t = \frac{\lambda}{4n}

Now we will put the values to the above formula to reach the answer

= \frac{480nm}{4\times 2.0}

= 60

Therefore we simply applied the above formula to determine the minimum coating thickness

6 0
3 years ago
The_______ of an object is given relative to an origin
Setler79 [48]
The slope of an object is given relative to an origin.
8 0
3 years ago
A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?
andriy [413]

Answer:

v = 22.54 mph.

Explanation:

Given that,

Distance moved, d = 200 m

Time, t = 19.8 s

We need to find the runner's average speed.

We know that,

1 mile = 1609.34 m

200 m = 0.124 miles

19.8 seconds = 0.0055 h

So,

Speed = distance/time

v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph

So, the runner's average speed is 22.54 mph.

4 0
2 years ago
If we cannot see the whole society what can we see
creativ13 [48]
I believe it's People interacting. I hope i've helped you! (:

8 0
2 years ago
the specific heat of water is 4.2 j/c. if it takes 31,500 joules to heat to warm 750 g of water, what was the temperature change
defon
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=mC_s \Delta T
where
m is the mass of the substance
C_s the specific heat capacity
\Delta T the increase in temperature

In our problem, the mass of the water is m=750 g, the specific heat is C_s = 4.2 J/g ^{\circ}C and the amount of heat supplied is Q=31500 J, so if we re-arrange the previous formula we find the increase in temperature of the water:
\Delta T= \frac{Q}{m C_s}= \frac{31500 J}{(750 g)(4.20 J/g^{\circ} C)}=10^{\circ}C
7 0
2 years ago
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