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marusya05 [52]
3 years ago
9

During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc

eleration? Show your work.
Physics
1 answer:
Lapatulllka [165]3 years ago
5 0
Let's calculate the average acceleration. It is the rate of changing speeds. Hence, we need to calculate the difference of speeds. 10-6=4 m/s. The rate is now \frac{4m/s}{2s} =2m/s^2.
In general, the formula for the mean acceleration between two times 1 and 2 is given by:
\frac{u_2-u_1}{T} where v1 and v2 are the speeds at the respective points and T is the time interval between them.
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How can accuracy be limited?
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Answer:

accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured

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3 years ago
Câu 1. Trường hợp nào dưới đây không phải là vật sáng?
Marianna [84]

Answer:

A

Explanation:

A. The pencil is on the table in broad daylight

5 0
3 years ago
An object is dropped from a height of 75.0 m above ground level. (a) Determine the distance traveled during the first second. (b
lys-0071 [83]

Answer:

a)Distance traveled during the first second = 4.905 m.

b)Final velocity at which the object hits the ground = 38.36 m/s

c)Distance traveled during the last second of motion before hitting the ground = 33.45 m

Explanation:

a) We have equation of motion

             S = ut + 0.5at²

     Here u = 0, and a = g

              S = 0.5gt²

    Distance traveled during the first second ( t =1 )

              S = 0.5 x 9.81 x 1² = 4.905 m

   Distance traveled during the first second = 4.905 m.

b)  We have equation of motion

            v² = u² + 2as

      Here u = 0, s= 75 m and a = g

           v² = 0² + 2 x g x 75 = 150 x 9.81

           v = 38.36 m/s

      Final velocity at which the object hits the ground = 38.36 m/s

c) We have S = 0.5gt²

                   75 = 0.5 x 9.81 x t²

                    t = 3.91 s

   We need to find distance traveled last second

   That is

          S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m

   Distance traveled during the last second of motion before hitting the ground = 33.45 m

       

3 0
3 years ago
What was significant about the discovery of gallium
Kryger [21]
It confirmed medeleeve's hypothesis (prediction) and showed the use of his table
8 0
3 years ago
Read 2 more answers
The speed of light in a transparent medium is 0.6 times that of its speed in vacuum. Find the refractive index of the medium.
grin007 [14]
<h2>Answer:</h2>

The refractive index is 1.66

<h2>Explanation:</h2>

The speed of light in a transparent medium is 0.6 times that of its speed in vacuum .

Refractive index of medium = speed of light in vacuum / speed of light in medium  

So

RI = 1/0.6 = 5/3 or 1.66

3 0
3 years ago
Read 2 more answers
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