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skad [1K]
3 years ago
12

In Michael Johnson's world-record 400 m sprint, he ran the first 100 m in 11.20 s; then he reached the 200 m mark after a total

time of 21.32 s had elapsed, reached the 300 m mark after 31.76 s,and finished in 43.18 s.
During what 100 m segment was his speed the highest?

a) Between the start and the 100 m mark.

b) Between the 100 m mark and the 200 m mark.

c) Between the 200 m mark and the 300 m mark.

d)Between the 300 m mark and the finish.
Physics
1 answer:
GrogVix [38]3 years ago
6 0

Answer:

b)

Explanation:

Assuming that we are talking about average speed during any segment, we can apply the definition of average speed, as follows:

v(avg) = Δx / Δt = (xf-x₀) / (tfi-t₀)

Using this definition for the 4 segments, we have:

1) v(0-100m) = 100 m / 11.20 sec = 8.93 m/s

2) v(100m-200m) = 100 m / (21.32 s - 11.2 s) = 100 m / 10.12 s = 9.88 m/s

3) v(200m -300m) = 100 m / (31.76 s- 21.32s) = 100 m / 10.44 s = 9.58 m/s

4) v(300m-400m) = 100 m / (43.18 s - 31.76 s) = 100 m / 11.42 s = 8.76 m/s

As we can see, the highest speed was reached between the 100m mark and the 200m mark, so the statement b) is the one that results to be true.

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Suppose you were hanging in empty space at rest, far from the Earth, but at the same distance from the Sun as the Earth. What mi
schepotkina [342]

Answer:

V = 42187 m/s = 42.18 km/s

Explanation:

given data:

mass of sun is  = 2\times 10^{30} kg

radius of earth orbit is 1.5\times 10^{11} m

minimum speed can be determined by using following formula

V = (\frac{2gM}{r}))^{1/2}

where G is \times 10^{-11}

Plugging all value to get desired value

V  =(\frac{2\times 6.674 \times 10^{-11}2\times 10^{30}}{1.5\times 10^{11}})^{1/2}

V = 42187 m/s = 42.18 km/s

4 0
2 years ago
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A and B, move toward one another. Object A has twice the mass and half the speed of object B. Which of the following describes t
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Answer:D

Explanation:

Given

mass of A is twice the mass of B half the velocity of B

Suppose F_a and F_b be the average force exerted on A and B respectively

and According to Newton third law of motion Force on the body A is equal to Force on body B but opposite in direction as they are action and reaction force.

Thus F_a=-F_b  and option d is correct

4 0
2 years ago
Vector A⃗ has magnitude 8.00 m and is in the xy-plane at an angle of 127∘ counterclockwise from the +x–axis (37∘ past the +y-axi
WINSTONCH [101]

Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i+ b j , where and a and b are constants to be found

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

- 12 j = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

Comparing coefficients of i and j:

a = 8 sin (37) = 4.81452 m

b = -12 - 8cos(37) = -18.38908

Hence,

B = 4.81452 i  - 18.38908 j  ..... 4 th quadrant

Hence,

cos ( Q ) = 4.81452 / 12

Q = 66.346 degrees

360 - Q = 293.65 degrees from + x-axis in CCW direction

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2 years ago
Compare the components that make up the windsurfer, his board, and his surroundings.
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Answer:

Sample Response: The windsurfer, his board, and the air and water around him are all made of matter. That matter is made up of very small particles called atoms.

Explanation:

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7 0
2 years ago
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A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

Given that,

Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

Electric field due to charge 2 is given by :

E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

So, the electric field strength at the midpoint between the two charges is 91406.24 N/C. Hence, this is the required solution.

3 0
2 years ago
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