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ella [17]
3 years ago
11

A bicyclist rides 2.93 km due east, while the resistive force from the air has a magnitude of 8.65 N and points due west. The ri

der then turns around and rides 2.93 km due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 8.65 N and points due east. Find the work done by the resistive force during the round trip. Number Type your answer here Units Choose your answer here
Physics
1 answer:
Ilya [14]3 years ago
4 0

Answer:

-50.6 kJ

Explanation:

The work done (W) on an object is given by:

W = (Fcosθ) * S

where F is the force, S is the displacement and θ is the angle between the force and displacement.

i) During the first trip riding east, S₁ = 2.93 km = 2930 m, F₁ = 8.65 N.

The displacement is due east and the force is due west, hence θ₁ = 180°. Therefore:

W₁ = (F₁ * cosθ₁)S₁ = (8,65 * cos(180))2930 = -25.3 kJ

ii) i) During the second trip riding west, S₂ = 2.93 km = 2930 m, F₂ = 8.65 N.

The displacement is due west and the force is due east, hence θ₂ = 180°. Therefore:

W₂ = (F₂ * cosθ₂)S₂ = (8,65 * cos(180))2930 = -25.3 kJ

work done by the resistive force during the round trip is:

W = W₁ + W₂ = -25.3 kJ + (-25.3 kJ) = -50.6 kJ

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What is the force required to accelerate a baseball with a mass of 0.145 kilograms and an acceleration of 35.00 meters/second sq
Paladinen [302]

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3 years ago
the distance a spring will stretch varies directly with how much weight is attached to the spring. If a spring stretches 11 inch
Paha777 [63]
Direct variation involves ration and proportions, so 

you need to set up the proportion:

<span>11 / 75 = x / 65
 
Cross multiplying:

75x = 11*65

x = (11*65)/75

Solving, we get x = 9.533, </span>

<span>which rounds off to 9.5

Therefore, the spring will stretch up to 9.5 inches with 65 attached.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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4 0
3 years ago
3
Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

KE = \frac{1}{2}mv^2

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

U_s = \frac{1}{2}kx^2

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2

We can simplify and solve for 'x'.

mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}

Plug in the givens and solve.

x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}

3 0
2 years ago
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