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Jlenok [28]
3 years ago
6

You set out to design a car that uses the energy stored in a flywheel consisting of a uniform 101-kg cylinder of radius r that h

as a maximum angular speed of 470 rev/s. the flywheel must deliver an average of 2.40 mj of energy for each kilometer of distance. find the smallest value of r for which the car can travel 300 km without the flywheel needing to be recharged.
Physics
1 answer:
Ket [755]3 years ago
8 0
Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is

E = (1/2)Iω²  

where I is the moment of inertia of the fly wheel and ω is the angular velocity.  
So this comes down to finding the value of I that gives the required energy.  We know the mass is 101kg.  The formula for a solid cylinder's moment of inertia is

 I = (1/2)mR²

We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec.  However, ω must be in radians per second so multiply it by 2π to get 
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω²,  for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²

Now find the radius R,

 165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m

R = 1.807m
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How much power is needed to lift a 20-kg object to a height of 4.0 m in 2.5 seconds?
4vir4ik [10]

Answer:

313.92w

Explanation:

Formula for power:

P=W/∆t = Fv

Givens:

m=20kg

∆y=4.0m

∆t=2.5s

a=9.81m/s²

In order to find power, we first need to solve for work.

W=Fd (force*displacement), f=mg

W=mg∆y

W=(20kg)(9.81m/s²)(4.0m)

W=784.8J

P=W/∆t

P=784.8J/2.5s

P=313.92 watts

5 0
2 years ago
Kissa and Lily ran a race in gym class Lily was off balance so she did not start until 2 seconds had passed
Paul [167]
Average speed = (distance covered) / (time to cover the distance)

Tissa covered 60 meters in 10 seconds. Her average speed was

(60 m) / (10 sec) = 6 m/s.

That's the slope of the dotted line.

Lilly covered 60 meters in 8 seconds. Her average speed was

(60 m) / (8 sec) = 7.5 m/s .

That's the slope of the solid line.

Lilly covered the same distance in less time, and both girls
arrived at the finish line together. Technically, in science talk,
we would say that Lilly ran "faster", and her average speed
was "greater".

We can detect that by looking at the graph, because Lilly's line
has the characteristic of being "steeper", and we know that the
slope of the line on a distance/time graph is "speed".
3 0
2 years ago
1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0
2 years ago
a horse runs with an initial velocity of 11m/s and slows to 5.2 m/s over a time interval of 3.1 s what is the horse's average ac
Katarina [22]

Answer:

a = change in v / change in time

= (5.2 - 11) / 3.1

= -1.87 m/s^2

Explanation:

6 0
3 years ago
If 1.00 mol of argon is placed in a 0.500-L container at 28.0 ∘C , what is the difference between the ideal pressure (as predict
Rudik [331]

Answer:

1.98 atm

Explanation:

Given that:

Temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28 + 273.15) K = 301.15 K

n = 1

V = 0.500 L

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L atm/ K mol  

Applying the equation as:

P × 0.500 L = 1 ×0.0821 L atm/ K mol  × 301.15 K

⇒P (ideal) = 49.45 atm

Using Van der Waal's equation

\left(P+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT

R = 0.0821 L atm/ K mol  

Where, a and b are constants.

For Ar, given that:

So, a = 1.345 atm L² / mol²

b =  0.03219 L / mol

So,  

\left(P+\frac{1.345\times \:1^2}{0.500^2}\right)\left(0.500-1\times 0.03219\right)=1\times 0.0821\times 301.15

P+\frac{1.345}{0.25}=\frac{24.724415}{0.46781}

P=\frac{24.724415}{0.46781}-\frac{1.345}{0.25}

⇒P  (real) = 47.47 atm

Difference in pressure = 49.45 atm - 47.47 atm = 1.98 atm

4 0
2 years ago
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