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VARVARA [1.3K]
3 years ago
12

Identify the position where the cyclist has maximum kinetic energy.

Physics
2 answers:
Gnom [1K]3 years ago
6 0

Answer:

the correct answer is the third position

finlep [7]3 years ago
3 0
You need the picture. However, since I'm doing PLATO as well, here's the answer:

The third position in which the cyclist is going downhill is the position where the cyclist has the maximum kinetic energy. The other two photos show that the biker has more potential energy.

Thanks, let me know if this helped!
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A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr
valina [46]
The  spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
T=2 t = 2 \cdot 0.100 s = 0.200 s
Which means that the frequency is
f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz
and the angular frequency is
\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s

In a spring-mass system, the maximum velocity of the object is given by
v_{max} = A \omega
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s
6 0
3 years ago
A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t
Natasha_Volkova [10]

Answer:

v=12.5 i + 12.5 j m/s

Explanation:

Given that

m₁=m₂ = m

m₃ = 2 m

Given that speed of the two pieces

u₁=- 25 j m/s

u₂ =- 25 i m/s

Lets take the speed of the third mass = v m/s

From linear momentum conservation

Pi= Pf

0 = m₁u₁+m₂u₂ + m₃ v

0 = -25 j m  - 25 i m + 2 m v

2 v=25 j   + 25 i m/s

v=12.5 i + 12.5 j m/s

Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s

4 0
3 years ago
Which term is used to describe the variety of inheritable traits in a species?
choli [55]
Genetic i think is it right?
7 0
3 years ago
Read 2 more answers
a grinding wheel is initially rotating with an angular velocity 5500 rad/srad/s when its motor is suddenly turned off. it comes
kiruha [24]

The angle through which the grinding wheel rotates in the first second =  <u>5300 rad</u>

Angular velocity is, the time charge at which an object rotates, or revolves, about an axis, or at which the angular displacement between our bodies changes. within the discern, this displacement is represented via the angle θ among a line on one body and a line on the alternative.

The angular velocity is described as the charge of trade of the angular position of a rotating body. Linear speed is defined because the charge of change of displacement with respect to time whilst the item moves alongside a straight course.

Initial angular velocity of the grinding wheel = ω1 = 5500 rad/s

Final angular velocity of the grinding wheel = ω2 = 0 rad/s   (Comes to rest)

Time is taken by the grinding wheel to come to rest = T = 10 sec

Angular acceleration of the grinding wheel = α

2 = ω1 + αT

0 = 5500 + α(10)

α = - 400 rad/s2

Negative as it is deceleration.

The angle through which the grinding wheel rotates in the first second = θ

Time period = T1 = 1 sec

θ = ω₁T1 + αT1²/2

θ = (5500)(1) + (-400)(1)²/2

θ = 5300 rad

The angle through which the grinding wheel rotates in the first second =  <u>5300 rad</u>

Learn more about angular velocity here:-brainly.com/question/6860269

#SPJ4

5 0
1 year ago
An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

so

= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

and now apply newton second law

force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
3 years ago
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