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Lelu [443]
3 years ago
14

A cyclist starts from rest and coasts down a 4.0∘ hill. The mass of the cyclist plus bicycle is 85 kg. Ignore air resistance and

friction.
A) After the cyclist has traveled 180 m , what was the net work done by gravity on the cyclist?

B) How fast is the cyclist going? in m/s

please explain if possible!!
Physics
1 answer:
Angelina_Jolie [31]3 years ago
5 0

A) change in ht after 180m = 180 * sin(4-deg.) = 12.56m

net work done by gravity on the cyclist = mass * gravity * height diff.

= 85 * 9.8 * 12.56

= 10470J

= 10.5kJ

B) Kinetic energy = 1/2 * mass * vel.^2 = work done by gravity = 10470J

vel.^2 = 10470 * 2 / 85 = 246.4

vel. = 15.7m/s

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In case A below, a 1 kg solid sphere is released from rest at point S. It rolls without slipping down the ramp shown, and is lau
mestny [16]

Answer:

the block reaches higher than the sphere

\frac{y_{sphere}} {y_block} = 5/7

Explanation:

We are going to solve this interesting problem

A) in this case a sphere rolls on the ramp, let's find the speed of the center of mass at the exit of the ramp

Let's use the concept of conservation of energy

starting point. At the top of the ramp

         Em₀ = U = m g y₁

final point. At the exit of the ramp

         Em_f = K + U = ½ m v² + ½ I w² + m g y₂

notice that we include the translational and rotational energy, we assume that the height of the exit ramp is y₂

energy is conserved

          Em₀ = Em_f

         m g y₁ = ½ m v² + ½ I w² + m g y₂

angular and linear velocity are related

        v = w r

the moment of inertia of a sphere is

         I = \frac{2}{5} m r²

we substitute

         m g (y₁ - y₂) = ½ m v² + ½ (\frac{2}{5} m r²) (\frac{v}{r})²

         m g h = ½ m v² (1 + \frac{2}{5})

where h is the difference in height between the two sides of the ramp

h = y₂ -y₁

         mg h = \frac{7}{5} (\frac{1}{2} m v²)

         v = √5/7  √2gh

This is the exit velocity of the vertical movement of the sphere

         v_sphere = 0.845 √2gh

B) is the same case, but for a box without friction

   starting point

          Em₀ = U = mg y₁

   final point

          Em_f = K + U = ½ m v² + m g y₂

          Em₀ = Em_f

          mg y₁ = ½ m v² + m g y₂

          m g (y₁ -y₂) = ½ m v²

          v = √2gh

this is the speed of the box

          v_box = √2gh

to know which body reaches higher in the air we can use the kinematic relations

          v² = v₀² - 2 g y

at the highest point v = 0

           y = vo₀²/ 2g

for the sphere

           y_sphere = 5/7 2gh / 2g

           y_esfera = 5/7 h

for the block

           y_block = 2gh / 2g

            y_block = h

       

therefore the block reaches higher than the sphere

         \frac{y_{sphere}} {y_bolck} = 5/7

3 0
3 years ago
Two spacemen are floating together with zero speed in a gravity-free region of space. The mass of spaceman A is 120 kg and that
maw [93]

Answer:

your answer is A

Explanation:

4 0
2 years ago
Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is
s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: \frac{∆v}{t}

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 \frac{feet}{second} then sped up to 5 \frac{feet}{second}, then the change in my speed would be 2 \frac{feet}{second}, because I started walking 2 \frac{feet}{second} faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 \frac{meters}{second} to 16 \frac{meters}{second}  in 4 seconds. What would the acceleration be? Let's set up an equation:

a = \frac{∆v}{t}

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 \frac{meters}{second} by 4 \frac{meters}{second}. That makes sense, right? Back to the equation.

a = \frac{∆v}{t}
a = \frac{16-4}{4}

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = \frac{12}{4}

a = 3 (\frac{meters}{second^{2}})

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit \frac{\frac{meters}{second}}{second}, which can be simplified down to \frac{meters}{second^{2} })

Let me know if you need clarification on anything I explained here!
- breezyツ

6 0
2 years ago
Plz help This is for Rock formations...only 5 sentences if not more
dsp73

Answer:

The coastal zone is not a stable and constant environment, but a dynamic place that can change rapidly in response to natural processes such as seasonal weather patterns. Waves, winds, currents, tides and storms are the major forces on the coast.

Explanation:

8 0
3 years ago
Hiii please help i’ll give brainliest if you give a correct answer please thanks!
mrs_skeptik [129]

Answer:

25 to the right

Explanation:

there you go friend your awsome

4 0
3 years ago
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