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barxatty [35]
3 years ago
15

The wavelength of green light is about the size of an atom. (T/F)

Physics
1 answer:
Snowcat [4.5K]3 years ago
5 0

Explanation:

The wavelength of green light is about 500 nanometers, or two thousandths of a millimeter. The typical wavelength of a microwave oven is about 12 centimeters, which is larger than a baseball.

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If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out
Maru [420]

Answer:

power=work done÷time taken

2×5=10

10÷10=1

ans 1J per second

5 0
2 years ago
Which of these experiments tests a chemical property of an object??
Ilia_Sergeevich [38]
<span>B. shining a bright light on the objects and testing for decomposition </span> <span>      

In explanation, chemical property is a characteristic of a certain substance came from an outcome due to chemical change or reaction. In the situation above, more specifically toxicity is involved in the chemical property/change. Hence, when the object is tested for decomposition. Like for an example of decomposition simply in metals, rusting. Rusting a process of degeneration of metals. Here it works the same. Toxicity is how much damage did a certain entity do to the object. </span>



8 0
3 years ago
Read 2 more answers
One litre of crude oil weighs 9.6N. Calculate its specific weight, density and specific gravity.​
Zepler [3.9K]

Answer:

The answer is "\bold{9600 \frac{N}{m^3}, 978.59 \frac{kg}{m^3}, and \ 0.978}"

Explanation:

Given:

\to v=1\ liter= 10^{-3} \ m^3\\\\\to  w= 9.6 \ N\\

calculation:

Specific \ weight =\frac{w}{v}=\frac{9.6}{10^{-3}}=9600 \frac{N}{m^3} \\\\w=mg\\\\m= \frac{w}{g}=\frac{9.6}{9.81}=0.9785\ kg\\\\\rho\ (density)=\frac{m}{v}=\frac{0.9785}{10^{-3}}=978.59 \frac{kg}{m^3}\\\\specific \ gravity = \frac{\prho \ obj}{\rho w}=\frac{978.54}{1000}=0.978

4 0
2 years ago
A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c
stealth61 [152]

Answer :

(a) The acceleration  of the car is, -2m/s^2

(b) The distance covered by the car is, 150 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 5 m/s

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = ?

Now put all the given values in the above equation 1, we get:

5m/s=25m/s+a\times (10s)

a=-2m/s^2

The acceleration  of the car is, -2m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 25 m/s

t = time = 10 s

a = acceleration  of the car = -2m/s^2

Now put all the given values in the above equation 2, we get:

s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2

By solving the term, we get:

s=150m

The distance covered by the car is, 150 m

8 0
3 years ago
4. What is the acceleration of the car in each section?<br> b<br> с<br> d<br> a
makvit [3.9K]

Answer:

0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.

Explanation:

3 0
2 years ago
Read 2 more answers
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