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lbvjy [14]
3 years ago
11

A runner is running to the left with a speed of 8.0\,\dfrac{\text m}{\text s}8.0 s m ​ 8, point, 0, start fraction, start text,

m, end text, divided by, start text, s, end text, end fraction when she speeds up with constant acceleration for 2.0\,\text s2.0s2, point, 0, start text, s, end text to reach the finish line. The runner's final speed is 9.9\,\dfrac{\text m}{\text s}9.9 s m ​ 9, point, 9, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the left. What was the runner's acceleration as she sped up to the finish line?
Physics
2 answers:
Alla [95]3 years ago
4 0

Answer:

The runner's acceleration as she sped up to the finish line is 0.95m/s²

Explanation:

Acceleration is the change in velocity of a body with respect to time. It is expressed as;

Acceleration = change in velocity/time

Change in velocity = final velocity - initial velocity

Acceleration = final velocity - initial velocity / time

Given initial speed = 8.0m/s

Final speed = 9.9m/s

Time taken = 2.0s

Acceleration = 9.9-8.0 / 2.0

Acceleration = 1.9/2

Acceleration = 0.95m/s²

yawa3891 [41]3 years ago
4 0

Answer:

-0.95

Explanation:

saw it on khan academy

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Answer:

Would love to be your friend m'lady.

Explanation:

8 0
3 years ago
A dog runs from point O to B. The dog reaches A at 2.3 s into the sprint. Then the dog reaches C at 4.1 s. What is the average v
anygoal [31]
It is 3.2 because you add 2.3 and 4.1 then divide your answer by 2.
7 0
3 years ago
A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At th
lana66690 [7]

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

7 0
2 years ago
Read 2 more answers
A negative charge, q1, of 6 µC is 0. 002 m north of a positive charge, q2, of 3 µC. What is the magnitude and direction of the e
prohojiy [21]

Force on the particle is defined as the application of the force field of one particle on another particle. The magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

<h3>What is electrical force?</h3>

Force on the particle is defined as the application of the force field of one particle on another particle. It is a type of virtual force.

The given data in the problem is

q₁ is the negative charge = 6 µC=6×10⁻⁶ C

q₂ is the positive charge = 3 µC=3×10⁻⁶ C

r is the distance between the charges=0.002 m

F_E is the electric force =?

The value of electric force will be;

\rm F_E= \frac{Kq_1q_2}{r^2} \\\\ F_E= \frac{9\times 10^9\times 6\times 10^{-6}\times3\times10^{-6}}{(0.002)^2}\\\\ \rm F_E=4.05\times10^4\;N

Hence the magnitude and direction of the electrical force will be 4.05×10⁴N towards the north.

To learn more about the electrical force refer to the link;

brainly.com/question/1076352

7 0
2 years ago
The two measurements necessary for calculating average speed are
alisha [4.7K]

The correct answer is Option (C) distance and time

Explanation:

Average speed of any object is defined as the total distance that object travels over the time it takes to travel that distance. In other words, average speed is the total distance divided by the elapsed time.

Average \thinspace Speed = \frac{Total \thinspace Distance}{Elapsed \thinspace Time}

Therefore, as you can see in the above equation, the two measurements that are essential for the calculation of the average speed are the (total) distance and the (elapsed) time.

Hence, the correct option is C.

5 0
3 years ago
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