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Klio2033 [76]
3 years ago
13

A student has a mass (including clothes and shoes) of 65.0 kg. She drinks a 12 oz. can of soda, with a nutritional energy conten

t of 140 Cal. Assuming that the efficiency of her muscles is 20%, how high can she climb without losing weight?
Conversion: 1 Cal = 1,000 cal = 4,180 J
Physics
1 answer:
Vikentia [17]3 years ago
4 0

She can climb 0.92 m without losing weight.

<u>Explanation:</u>

Gravitational potential energy is the energy consisting of the product of mass, gravity and height.

1 cal = 4184 J

140 cal = 585760 J

Energy = 585760 J,  m = 65.0 kg = 65000 g,   Efficiency = 20 %

                                  GPE = mgh

where m represents the mass

           g represents the gravity,

           h represents the height.

                             585760 = 65000  9.8  h

                                        h = 0.92 m.

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Assuming a vertical trajectory with no drag, derive the applicable form of the rocket equation for this application
VARVARA [1.3K]

Answer:

The vertical trajectory is governed by Ordinary Differential Equation.

Time derivatives of each state variables.

d(d)/dt = v, d(m)/dt = -d(m-fuel)/dt, d(v)/dt = F/m.

Where V is velocity positive upwards, t is time, m is mass, m-fuel is fuel mass, F is Total force, positive upwards.

Therefore,

F = -mg - D + T, If V is positive and

F = -mg + D - T, If T is negative.

D is drag and the questions gave it as zero.

Explanation:

The two sign cases in derivative equations above are required because F is defined positive up, so the drag D and thrust T can subtract or add to F depending in the sign of V . In contrast, the gravity force contribution mg is always negative. In general, F will be some function of time, and may also depend on the characteristics of the particular rocket. For example, the T component of F will become zero after all the fuel is expended, after which point the rocket will be ballistic, with only the gravity force and the aerodynamic drag force being p

8 0
4 years ago
he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?
Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}

With the following condition :

  • \sf{\omega} = angular frequency (rad/s)
  • \sf{\theta} = change of angle value (rad)
  • t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

  • \sf{\theta} = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
  • t = interval of the time = 54.9 s.

What was asked :

  • \sf{\omega} = angular frequency = ... rad/s

Step by step :

\sf{\omega = \frac{\theta}{t}}

\sf{\omega = \frac{2,000 \pi}{54.9}}

\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

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