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ella [17]
2 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]2 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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The parallel plates in a capacitor, with a plate area of 7.90 cm2 and an air-filled separation of 2.70 mm, are charged by a 7.90
s2008m [1.1K]

Answer:

A) 26V

Explanation:

(a) the potential difference between the plates

Initial capacitance can be calculated using below expresion

C1= A ε0/ d1

Where d1= distance between = 2.70 mm= 2.70× 10^-3 m

ε0= permittivity of space= 8.85× 10^-12 Fm^-1

A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2

If we substitute the values we

C1= A ε0/ d1

=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3

C1=2.589 ×10^-12 F= 2.59 pF

Initial charge can be determined using below expresion

q1= C1 × V1

V1=2.589 ×10^-12 F

V1= voltage=7.90 V

If we substitute we have

q1= 2.589 ×10^-12 × 7.90

q1= 20.45×10^-12C

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C2= A ε0/ d2

d2=8.80 mm= /8.80× 10^-3

7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3

C1=0.794 ×10^-12 F= 0.794 pF

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q2=q1= 20.45 pC

Final potential difference

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= 20.45/0.794

= 26V

6 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

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z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

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\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

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150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

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5 0
3 years ago
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True.

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