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goblinko
2 months ago
10

A bicycle rider has a speed of 19.0 m/s at a height of 55.0 m above sea level when he begins coasting down hill. The mass of the

rider and his bike is 88.0 kg. Sea level is the zero level for measuring gravitational potential energy. Ignoring friction and air resistance, what is the rider’s total mechanical energy when he coasts to a height of 25.0 m above sea level
Physics
1 answer:
ValentinkaMS [3.4K]2 months ago
5 0
The total mechanical energy of the rider at any height amounts to 6.34 × 10⁴ J. The rider's mechanical energy is computed as the total of gravitational potential energy and kinetic energy. Assuming no forces are lost (like friction), this mechanical energy remains constant at different heights, as potential energy lost translates into kinetic energy gained, in accordance with the conservation of energy principle. Calculating both potential and kinetic energy at 55.0 m and 19 m/s allows us to derive the consistent mechanical energy: Mechanical energy = PE + KE Where: PE = potential energy, KE = kinetic energy. The potential energy calculation goes as follows: PE = m · g · h, Where: m is the mass, g is the gravitational acceleration, h represents the height. The calculation of the rider's potential energy yields: PE = 88.0 kg · 9.81 m/s² · 55.0 m = 4.75 × 10⁴ J. The kinetic energy is derived from: KE = 1/2 · m · v², Here "m" refers to the mass and "v" stands for velocity. Thus, KE = 1/2 · 88.0 kg · (19.0 m/s)² = 1.59 × 10⁴ J. Consequently, the rider's mechanical energy calculates as: Mechanical energy = PE + KE = 4.75 × 10⁴ J + 1.59 × 10⁴ J = 6.34 × 10⁴ J. This mechanical energy remains unchanged since when the rider descends, potential energy is converted into kinetic energy, keeping the total sum of these energies constant.
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Sav [3153]

mass₃<mass₁=mass₅<mass₂=mass₄

Explanation:

Data points:-

1. mass:  m      speed: v

2. mass: 4 m   speed: v

3. mass: 2 m   speed: ¼ v

4. mass: 4 m   speed: v

5. mass: 4 m   speed: ½ v

We know that the formula for Kinetic energy (KE) is ½ mv²

Where m represents the mass of the object

           v represents the object's velocity

<psubstituting the="" given="" values="" for="" mass="" and="" speed="" from="" previous="" data:="">

The KE of Body 1(mass₁) = ½*m*v²             = mv²/2

KE of Body 2(mass₂) = ½*4m*v²         = 2mv²

KE of Body 3(mass₃) = ½*2m*(1/4v)²  =  mv²/16

KE of Body 4(mass₄) = ½*4m*v ²        =  2mv ²

KE of Body 5(mass₅) = ½*4m*(1/2v)²  =   mv²/2

</psubstituting>
6 0
2 months ago
Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
Maru [3345]

Answer:

Jari

Explanation:

To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.

For Jari's path, starting point is (0, 0) and (6, 7) is another point.

The gradient is the difference in y divided by the difference in x:

Change in y=7-0=7

Change in x=6-0=6

Thus, the slope equals 7/6.

For Jade, her first point is (0, 10) and another is (6, 16).

Change in y=16-10=6

Change in x=6-0=6

Thus, the slope equals 6/6=1.

It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.

3 0
2 months ago
Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of t
Softa [3030]

Answer:

1/7 kg

Explanation:

Refer to the attached diagram for enhanced clarity regarding the question.

One of the blocks weighs 1.0 kg and accelerates downward at 3/4g.

g denotes the acceleration due to gravity.

Let M represent the block with known mass, while 'm' signifies the mass of the other block and 'a' refers to the acceleration of body M.

Given M = 1.0 kg and a = 3/4g.

By applying Newton's second law; \sum fy = ma_y

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Mg - T = Ma... (2)

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M(a-g) = -m(a+g)

Substituting M = 1.0 kg and a = 3/4g into this equation leads to;

3/4 g-g = -m(3/4 g+g)

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Keith_Richards [3271]

Result:

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ValentinkaMS [3465]

Response: The result would be no reaction.

Clarification:

7 0
2 months ago
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