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grin007 [14]
2 years ago
9

Define light?it's types​

Physics
1 answer:
sineoko [7]2 years ago
3 0

Answer:

Light or visible light is electromagnetic radiation within the portion of the electromagnetic spectrum that can be perceived by the human eye. Visible light is usually defined as having wavelengths in the range of 400–700 nm, or 4.00 × 10⁻⁷ to 7.00 × 10⁻⁷ m, between the infrared and the ultraviolet.

Explanation:

This is one type of light.

You might be interested in
How do I solve for time if I have initial velocity and final speed?
xeze [42]
Final speed = initial speed + (acceleration x time)

(final speed - initial speed) = acceleration x time

Time = (final speed - initial speed) / acceleration
4 0
3 years ago
6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
A force of 5 n produces an acceleration of 8m/s2 in mass m1 and an acceleration of 24 m/s2 in mass m2 .what acceleration would i
Inessa [10]

Acceleration of the both masses tied together= 6m/s²

Explanation:

The force is given by F= ma

so 5= m1 (8)

m1=0.625 Kg

for m2

5=m2 (24)

m2=0.208 kg

now total mass= m1+m2=0.625+0.208

Total mass=M=0.833 Kg

now F= ma

5= 0.833 (a)

a= 5/0.833

a=6m/s²

4 0
3 years ago
A car, starting from the origin, travels
sweet [91]

Answer:

The net displacement of the car is 3 km West

Explanation:

Please see the attached drawing to understand the car's trajectory: First in the East direction for 4 km (indicated by the green arrow that starts at the origin (zero), and stops at position 4 on the right (East).

Then from that position, it moves back towards the West going over its initial path, it goes through the origin and continues for 3 more km completing a moving to the West a total of 7 km. This is indicated in the drawing with an orange trace that end in position 3 to the left (West) of zero.

So, its NET displacement considered from the point of departure (origin at zero) to the final point where the trip ended, is 3 km to the west.

8 0
3 years ago
Read 2 more answers
Signment
jeka94

Answer:

v = 5.15 m/s

Explanation:

At constant velocity, the cable tension will equal the car weight of 984(9.81) = 9,653 N

As the cable tension is less than this value, the car must be accelerating downward.

7730 = 984(9.81 - a)

a = 1.95 m/s²

kinematic equations s = ut + ½at² and v = u + at

-5.00 = u(4.00) + ½(-1.95)4.00²

u = 2.65 m/s    the car's initial velocity was upward at 2.65 m/s

v = 2.65 + (-1.95)(4.00)

v = -5.15 m/s

3 0
2 years ago
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