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noname [10]
3 years ago
7

A girl on a bike is moving at a speed of 1.40 m/s at the start of a 2.45 m high and 12.4 m long incline. The total mass is 60.0

kg, air resistance and rolling resistance can be modeled as a constant friction force of 41.0 N, and the speed at the lower end of the incline is 6.70 m/s. Determine the work done (in J) by the girl as the bike travels down the incline.
Physics
1 answer:
Scrat [10]3 years ago
8 0

Answer:

Explanation:

Given that,

Initial speed of the girl is

u = 1.4m/s

Height she is going is

H = 2.45m

Incline plane she will pass to that height

L = 12.4m

Mass of girl and bicycle is

M=60kg

Frictional force that oppose motion is

Fr = 41N

Speed at lower end of inclined plane

V2 = 6.7m/s

Work done by the girl when the car travel downward

Using conservation of energy

K.E(top) + P.E(top) + work = K.E(bottom) + P.E(bottom) + Wfr

Where Wfr is work done by friction

Wfr = Fr × d

P.E(bottom) is zero, sicne the height is zero at the ground

K.E is given as ½mv²

Then,

½M•u² + MgH + W = ½M•V2² + 0 + Fr×d

½ × 60 × 1.4² + 60×9.8 × 2.45 + W = ½ × 60 × 6.7² + 41 × 12.4

58.8 + 1440.5 + W = 1855.1

W = 1885.1 —58.8 —1440.5

W = 355.8 J

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Setler [38]

Answer:

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Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

a = Acceleration

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The acceleration of the bicycle and rider is -1.5 m/s²

Force

F=ma\\\Rightarrow F=80\times -1.5\\\Rightarrow F=-120\ N=-1.2\times 10^2\ N

The magnitude of the average force needed to bring the bicycle and its rider to a stop is 1.2\times 10^2\ N

3 0
3 years ago
the positive particle has a charge of 31.7 mC and the particles are 2.80 mm apart, what is the electric field at point A located
vichka [17]

Answer:

the electric field at point A is

E = 5.5 ×10¹³N/C(-x direction)

Explanation:

given

electrostatics constant k = 9.0×10⁹

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distance r = 2.80mm

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attached is the diagram of the solution, describing the position of the charge

note x = r/2, where x is the distance from midpoint of r to the particle

using Pythagoras theorem as in the attachment, x = 2.44mm= 2.44×10⁻³m

the electric field at point A is given as

vector <em>E </em>= 2E×cos θ( -x direction)

recall E =kq/x²

where k is the electrostatics constant = 1/4πε₀

where ε₀ is permittivity of free space

therefore using E =2{kq/x²}cosθ

∴cosθ = adjacent/hypotenuse

cosθ=1.40/2.44

E =2 {(9.0×10⁹ × 31.7×10⁻³) ÷ (2.44×10⁻³)²}×(1.40/2.44)(-x)

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3 0
3 years ago
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MrRa [10]
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\Delta U = Q-W
where
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Using this form, the sign convention for Q and W becomes:
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Answer:

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explanation:

I hope it will help you

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Answer:

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