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Using energy considerations, calculate the average force (in N) a 67.0 kg sprinter exerts backward on the track to accelerate fr

Illusion [34]

Answer:

$F_{sprinter}=110.4N$

Explanation:

Given data

Mass m=67.0 kg

Final Speed vf=8.00 m/s

Initial Speed vi=2.00 m/s

Distance d=25.0 m

Force F=30.0 N

From work-energy theorem we know that the work done equals the change in kinetic energy

W=ΔK=Kf-Ki=1/2mvf²-1/2mvi²

And

$W=F_{total}.d$

So

$W=1/2mv_{f}^2-1/2mv_{i}^2\\F_{total}=\frac{1/2mv_{f}^2-1/2mv_{i}^2}{d} \\F_{total}=\frac{1/2(67.0kg)(8.00m/s)^2-1/2(67.0kg)(2.00m/s)^2}{25.0m} \\F_{total}=80.4N$

and we know that the force the sprinter exerted Fsprinter the force of the headwind Fwind=30.0N

So

$F_{sprinter}=F_{total}+F_{wind}\\F_{sprinter}=80.4N+30N\\F_{sprinter}=110.4N$

7 0

2 years ago

Read 2 more answers

A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down

Leona [35]

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

$wsin\theta-\mu F_n=ma$ That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

6 0

3 years ago

José and Laurel measured the length of a stick's shadow during the day. Without knowing the length of the stick, which of their

skelet666 [1.2K]

Answer:

Option B.

Explanation:

Assuming the stick is in vertical position, its shadow depends on two factors: its length and the angle between the sun rays and the stick. When the angle is bigger, the lenght of the shadow increases, and vice versa. So, when the sun rays are parallel to the stick, the shadow may be small. Since they are nearly perpendicular to the Earth's surface at 12 o'clock, the shadow of the stick at that time should be minimal. It means that the measured shadow of 75 cm at 12:30 p.m. is almost impossible (Option B).

5 0

2 years ago

What are the factor that affect gravity

valentinak56 [21]

Explanation:

The factors that affect gravity are as follows:

1. mass of body

2. acceleration

Keep smiling and hope u are satisfied with my answer.Have a good day :)

6 0

2 years ago

A 5.00X10^5 kg rocket is accelerating straight up. Its engines produce 1.250X10^7 N of thrust, and air resistance is 4.50X10^6 N

katrin [286]

6.20 m/s^2 The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be 9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N Add in the atmospheric drag and you get 4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N Now subtract that total drag from the thrust available. 1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So 3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2 Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2

8 0

3 years ago