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Digiron [165]
3 years ago
10

Use these values: distance traveled is 30.0 km; time elapsed is 40.0 minutes; average power output is 370 watts. Determine the t

otal energy expended by the cyclist, in kilojoules.
Physics
1 answer:
serious [3.7K]3 years ago
4 0

Explanation:

It is known that the relation between power and work is as follows.

           Power = \frac{\text{work done}}{Time}

As it is given that power is 370 watts and time is 40 minutes. First, we will convert minutes into seconds as follows.

               40 min \times \frac{60 sec}{1 min}

                 = 2400 sec

Now, total power or energy is calculated as follows.

              Power = \frac{\text{work done}}{Time}

                         = \frac{370 watts}{2400 sec}        

                         = 888000 J

As 1 kJ = 1000 J. So, convert 888000 J into kJ as follows.

                 \frac{888000}{1000} kJ

                  = 888 kJ

therefore, we can conclude that total energy expended by the cyclist is 888 kJ.

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Why does jupiter have several distinct cloud layers?
pentagon [3]

Different layers represent clouds made of gases that condense at different temperatures.

4 0
3 years ago
1. A soccer ball is kicked horizontally off a cliff with an initial speed of 8 m/s and lands 16 m from the base of
lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0
3 years ago
A 60 kg man jumps down from a 0.8 m table. What is the speed when he
Rom4ik [11]

Answer: Speed = 4 m/s

Explanation:

The parameters given are

Mass M = 60 kg

Height h = 0.8 m

Acceleration due to gravity g= 10 m/s2

Before the man jumps, he will be experiencing potential energy at the top of the table.

P.E = mgh

Substitute all the parameters into the formula

P.E = 60 × 9.8 × 0.8

P.E = 470.4 J

As he jumped from the table and hit the ground, the whole P.E will be converted to kinetic energy according to conservative of energy.

When hitting the ground,

K.E = P.E

Where K.E = 1/2mv^2

Substitute m and 470.4 into the formula

470.4 = 1/2 × 60 × V^2

V^2 = 470.4/30

V^2 = 15.68

V = square root (15.68)

V = 3.959 m/s

Therefore, the speed of the man when hitting the ground is approximately 4 m/s

4 0
3 years ago
Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi
Bad White [126]
A) Agreed. 
<span>b) Value agreed but units should be W (watts). </span>

<span>c) Here's one method... </span>

<span>15 miles = 24140 m </span>

<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>

<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>

<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>

<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>

<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>

<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
8 0
3 years ago
An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this
galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

\gamma = \frac{Mt}{J}

Where:

γ: angular acceleration

Mt: torque

J: moment of inertia of the load from its turning axis

Since we have the torque we just need the moment of inertia. We have to add together the moments of the drive shaft, tires, wheel walls and wheels.

The wheels act like disks. For disks the moment of inertia is:

J = \frac{1}{2} * m * r^2

Jwheel = \frac{1}{2} = 15 * 0.18^2 = 0.243 kg*m^2

The wheel walls act like annular rings, for these the moment of inertia is:

J = \frac{1}{2} * m * (re^2 - ri^2)

Jwall = \frac{1}{2} * 2 * (0.32^2 - 0.18^2) = 0.07 kg * m^2

The tread acts like a hoop, as in mass concentrated into a circunference, for these:

J = m * r^2

Jtread = 10 * 0.33^2 = 1.09 kg*m^2

The axle acts like a rod, which is the same as the disk:

Jaxle = \frac{1}{2} * 14.1 * 0.02^2 = 0.0028 kg*m^2

The drive shaft acts like a rod too:

Jshaft = \frac{1}{2} * 31.7 * 0.032^2 = 0.016 kg*m^2

SO, the total moment of inertia is:

J = 2*Jwheel + 2*Jwall + 2*Jtread + Jaxle + Jshaft

J = 2*0.243 + 2*0.07 + 2*1.09 + 0.0028 + 0.016 = 2.82 kg*m2

Finally the angular acceleration is:

\gamma = \frac{0.852 * 153}{2.82} = 46.2 \frac{rad}{s^2}

4 0
3 years ago
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