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An isothermal process is a change of a system, in which the temperature remains constant: ΔT = 0. This typically occurs when a system is in contact with an ... more
Escape velocity is the speed at which the kinetic energy plus the gravitational potential energy of an object is zero. It is the speed needed to ... more
The Euler’s pump and turbine equations are most fundamental equations in the field of turbo-machinery. These equations govern the power, efficiencies and ... more
Deformation in continuum mechanics is the transformation of a body from a reference configuration to a current configuration.
Strain is a normalized
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The Penman equation describes evaporation (E) from an open water surface, and was developed by Howard Penman in 1948. Penman’s equation requires ... more
A sled experiences a rocket thrust that accelerates it to the right.Each rocket creates an identical thrust T . As in other situations where there is only horizontal acceleration, the vertical forces cancel. The ground exerts an upward force N on the system that is equal in magnitude and opposite in direction to its weight,w.The system here is the sled, its rockets, and rider, so none of the forces between these objects are considered. The arrow representing friction ( f ) is drawn larger than scale.
Assumptions: The mass of the Sled remains steady throughout the operation
Strategy
Although there are forces acting vertically and horizontally, we assume the vertical forces cancel since there is no vertical acceleration. This leaves us with only horizontal forces and a simpler one-dimensional problem. Directions are indicated with plus or minus signs, with right taken as the positive direction. See the free-body diagram in the figure.
Solution
Since acceleration, mass, and the force of friction are given, we start with Newton’s second law and look for ways to find the thrust of the engines. Since we have defined the direction of the force and acceleration as acting “to the right,” we need to consider only the magnitudes of these quantities in the calculations. Hence we begin with
Fnet is the net force along the horizontal direction, m is the rocket’s mass and a the acceleration.
We can see from the Figure at the top, that the engine thrusts add, while friction opposes the thrust.
Tt is the total thrust from the 4 rockets, Fnet the net force along the horizontal direction and Ff the force of friction.
Finally, since there are 4 rockets, we calculate the thrust that each one provides:
T is the individual Thrust of each engine, b is the number of rocket engines
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/
The main span of San Francisco’s Golden Gate Bridge is 1275 m long at its coldest. The bridge is exposed to temperatures ranging from –15ºC to 40ºC . (a) What is its change in length between these temperatures? Assume that the bridge is made entirely of steel.
Strategy
Use the equation for linear thermal expansion to calculate the change in length , ΔL . Use the coefficient of linear expansion, α ,for steel from Table 13.2, and note that the change in temperature, ΔT , is 55ºC
(b) convert the change in temperature if Kelvin and Fahrenheit degrees. **
**this section is not included in the Reference material
Discussion
Although not large compared with the length of the bridge, this change in length is observable. It is generally spread over many expansion joints so that the expansion at each joint is small.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/
A typical small rescue helicopter, like the one in the Figure below, has four blades, each is 4.00 m long and has a mass of 50.0 kg. The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades. (c) To what height could the helicopter be raised if all of the rotational kinetic energy could be used to lift it?
The first image shows how helicopters store large amounts of rotational kinetic energy in their blades. This energy must be put into the blades before takeoff and maintained until the end of the flight. The engines do not have enough power to simultaneously provide lift and put significant rotational energy into the blades.
The second image shows a helicopter from the Auckland Westpac Rescue Helicopter Service. Over 50,000 lives have been saved since its operations beginning in 1973. Here, a water rescue operation is shown. (credit: 111 Emergency, Flickr)
Strategy
Rotational and translational kinetic energies can be calculated from their definitions. The last part of the problem relates to the idea that energy can change form, in this case from rotational kinetic energy to gravitational potential energy.
Solution for (a)
We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find Er . The angular velocity ω for 1 r.p.m is
and for 300 r.p.m
The moment of inertia of one blade will be that of a thin rod rotated about its end.
The total I is four times this moment of inertia, because there are four blades. Thus,
and so The rotational kinetic energy is
Solution for (b)
Translational kinetic energy is defined as
To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is
Solution for (c)
At the maximum height, all rotational kinetic energy will have been converted to gravitational energy. To find this height, we equate those two energies:
Discussion
The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades—something you probably would not suspect. The 53.7 m height to which the helicopter could be raised with the rotational kinetic energy is also impressive, again emphasizing the amount of rotational kinetic energy in the blades.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/
Find the terminal velocity of an 85-kg skydiver falling in a spread-eagle position.
where Vt is the terminal velocity, m is the mass of the skydiver, g is the acceleration due to gravity, Cd is the drag coefficient, ρ is the density of the fluid through which the object is falling, and A is the projected area of the object.
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/
where h is skydiver height and w the width at “spread-eagle” position
What is the cost of running a 0.600-kW computer 6.00 h per day for 30.0 d if the cost of electricity is $0.120 per kW ⋅ h ?
where t is the total consumption time, td is the days of consumption and th the hours of consumption per day
where P is Power consumption rate, E is the energy supplied by the electricity company and t is consumption time
keywords:
ballistics
where C is the total cost and CkW is the cost per kilowatt hour
Reference : OpenStax College,College Physics. OpenStax College. 21 June 2012.
http://openstaxcollege.org/textbooks/college-physics
Creative Commons License : http://creativecommons.org/licenses/by/3.0/
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Prior to manned space flights, rocket sleds were used to test aircraft, missile equipment, and physiological effects on human subjects at high speeds. They consisted of a platform that was mounted on one or two rails and propelled by several rockets. Calculate the magnitude of force exerted by each rocket, called its thrust T , for the four-rocket propulsion system shown in the Figure below. The sled’s initial acceleration is 49 m/s 2, the mass of the system is 2100 kg, and the force of friction opposing the motion is known to be 650 N.