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lord [1]
3 years ago
5

A car traveling on a straight road at 10 meters per second accelerates uniformly to a speed of 21 meters per second in 12 second

s. the total distance traveled by the car in this 12 seconds time interval is
Physics
1 answer:
labwork [276]3 years ago
6 0
21+10=31 because you can see that 21 and 10 are in metres while 12 is in seconds so 21+10=31 is the answer.
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Which wavelength of light is the best choice when attempting to quantitatively relate solution absorbance and concentration?.
ANEK [815]

The optimum wavelength is 450 nm because that is the wavelength of maximum absorbance by FeSCN2+(aq)

you should choose a wavelength with maximum absorbance. In this case, you are using the scattered light, not the absorbed light as your signal. So you should avoid wavelengths where there are absorption peaks.

<h3>What is wavelength ?</h3>

A waveform signal that is carried in space or down a wire has a wavelength, which is the separation between two identical places (adjacent crests) in the consecutive cycles. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm) (mm).

  • The distance between two waves' crests serves as an illustration of wavelength. When you and another person have the same overall mindset and can easily communicate, that is an example of being on the same wavelength.

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6 0
2 years ago
On a trolley ride around an amusement park, a child travelled from one signpost to a second signpost at a constant speed of 125
Tomtit [17]

Answer:

Explanation:

Speed given = 125 m /min

125 /60 m /s

In 450 second it will travel

= 450 x 125 / 60

=937.5 m.  

As the distance  is covered in less than 450 seconds , The distance must be less than 937.5 m

In 400 seconds , it will travel

= 400 x 125 / 60

833.33 m

Since the distance is covered in more than 400 seconds , the distance must be more than ie 833.33 .

Hence the distance covered is more than .833 m but less than 937.5

In either case these distance are more than .8 km .

5 0
3 years ago
Solve the problem.
gulaghasi [49]
As the shock waves travel in concentric outward circles from the epicenter, and the diameter is measured 120 miles,
area of a circle =<span>π</span><span>r*r</span>

d=120
<span>r=<span>120/2</span></span><span>r=60</span><span><span>60*60</span>=3600</span><span>3600*π=11309.734</span>
<span>11309.734 square miles</span>
5 0
3 years ago
What was the largest group in the colonies in the year leading up to the American revoulation
Degger [83]
Answer: The English origin was the largest

7 0
2 years ago
Read 2 more answers
A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
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