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trapecia [35]
2 years ago
13

An athlete is running a 400m race around a 400m track. On the backstretch the athlete's velocity is 8m/s but he is running into

a 2m/s headwind. How large is the drag force that acts on him? Assume that the density of the air is 1.2kg/m^3, his cross sectional area is 0.5m^2, and the coefficient of drag is 1.1.
Physics
1 answer:
Aleksandr-060686 [28]2 years ago
3 0

Answer:

33 N

Explanation:

v = Velocity of fluid = 8+2 = 10 m/s

\rho = Density of fluid = 1.2 kg/m³

C = Coefficient of drag = 1.1

A = Cross sectional area = 0.5 m²

Drag force is given by

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2\times 1.1\times 0.5\times (8+2)^2\\\Rightarrow F=33\ N

The drag force on the athlete is 33 N

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a 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they
Otrada [13]

According to the given statement Final velocity when they stick together is 8.735i^ + 11.25j^​

<h3>What is collision and momentum?</h3>

The unit of momentum is kg ms -1. Momentum is a vector parameter that is influenced by the object's direction. During collisions involving objects, momentum is a relevant concept. The final velocity before a collision between two objects equals the total motion after the impact (in the absence of external forces).

<h3>Briefing:</h3>

From conservation of momentum

Initial momentum = final momentum

m u +M U =(m+M) V

2000×25 i^ +1500×30 j^​ =(2000+1500) V

V = 8.735i^ + 11.25j^​

Final velocity when they stick together is 8.735i^ + 11.25j^​

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The complete question is -

A 2000 kg truck is moving eastward at 25 m/s. it collides inelastically with a 1500 kg truck traveling southward at 30 m/s. they collide at the intersection. Find the direction and magnitude of velocity of the wreckage after the collision, assuming the vehicles stick together after the collision.

6 0
1 year ago
three girls were pushing the same car with a net force of 450 N [N48°E]. Two of the girls were pushing with forces of 310 N [N25
ElenaW [278]

The net force is the vector

∑ F = (450 N) (cos(42°) i + sin(42°) j)

and two of the forces provided by the girls are

F₁ = (310 N) (cos(115°) i + sin(115°) j)

F₂ = (250 N) (cos(285°) i + sin(285°) j)

Then the force provided by the third girl is the vector

F₃ = ∑ F - F₁ - F₂

F₃ = ((450 N) cos(42°) - (310 N) cos(115°) - (250 N) cos(285°)) i

… … … + ((450 N) sin(42°) - (310 N) sin(115°) - (250 N) sin(285°)) j

F₃ ≈ (400.722 N) i + (261.635 N) j

So, the third girl provided a force of magnitude

||F₃|| = √((400.722 N)² + (261.635 N)²) ≈ 478.572 N ≈ 480 N

pointing in a direction

arctan((261.635 N)/(400.722 N)) ≈ 33.1409° ≈ 33°

relative to East which refers to 0°; that is, 33° N of E or E33°N. Since the other forces are given relative to North or South, we can write this direction as N57°E.

So, the third girl pushed with force 480 N [N57°E].

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