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2 years ago

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A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav

ity is g = 5.48 ft/s^2. Determine how much he will weigh (a) on the spring scale and (b) on the beam scale.The acceleration of high-speed aircraft is sometimes expressed in g’s (in multiples of the standard acceleration of gravity). Determine the net upward force, in N, that a 90-kg man would experience in an aircraft whose acceleration is 6 g’s.

1 answer:

kozerog [31]2 years ago

3 0

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

$g = 5.48 ft/s^2.$

m= 150 lbm

a)

Weight on the spring scale(Wt) = m g

We know that

$1\ lbf=32.17 \ lmb.ft/s^2$

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

b)

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

$1\ ft/s^2= 0.304\ m/s^2$

$5.48\ ft/s^2= 1.66\ m/s^2$

a=6 g's

$a=9.99\ m/s^2$

So

F = 90 x 9.99 N

F= 899.59 N

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Answer:

$q_{in} = 528.6\,\frac{kJ}{kg}$

Explanation:

Let assume that heating process occurs at constant pressure, the phenomenon is modelled by the use of the First Law of Thermodynamics:

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2 years ago

A spring-loaded toy gun is used to shoot a ball of mass m = 1.50 kg straight up in the air. The spring has spring constant k = 6

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Answer:

1) a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

2) The muzzle velocity of the ball is approximately 5.272 meters per second.

3) The maximum height of the ball is 1.417 meters.

Explanation:

1) Which of the following statements are true?

a) Mechanical energy is conserved because no dissipative forces perform work on the ball.

True, statement indicates that there is no air resistence and no friction between ball and the inside of the gun because the first never touches the latter one.

b) The forces of gravity and the spring have potential energies associated with them.

False, force of gravity do work on the ball and spring receives a potential energy at being deformated by the ball.

c) No conservative forces act in this problem after the ball is released from the spring gun.

False, the absence of no conservative forces is guaranteed for the entire system according to the statement of the problem.

2) According to the statement, we understand that spring is deformed and once released and just after reaching its equilibrium position, the muzzle velocity is reached. As spring deformation is too small in comparison with height, we can neglect changes in gravitational potential energy. By Principle of Energy Conservation, we describe the motion of the ball by the following expression:

$U_{k, 1}+K_{1}=U_{k,2}+K_{2}$ (Eq. 1)

Where:

$U_{k,1}$ , $U_{k,2}$ - Initial and final elastic potential energies of spring, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After using definitions of elastic potential and translational kinetic energies, we expand the equation above as:

$\frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2}) = \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})$

And the final velocity is cleared:

$m\cdot (v_{2}^{2}-v_{1}^{2}) = k\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2}-v_{1}^{2} =\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2}^{2} =v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2})$

$v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }$ (Eq. 2)

Where:

$v_{1}$ , $v_{2}$ - Initial and final velocities of the ball, measured in meters per second.

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass of the ball, measured in kilograms.

$x_{1}$ , $x_{2}$ - Initial and final position of spring, measured in meters.

If we know that $v_{1} = 0\,\frac{m}{s}$ , $k = 667\,\frac{N}{m}$ , $m = 1.50\,kg$ , $x_{1} = -0.25\,m$ and $x_{2} = 0\,cm$ , the muzzle velocity of the ball is:

$v_{2} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+\left(\frac{667\,\frac{N}{m} }{1.50\,kg} \right)\cdot [(-0.25\,m)^{2}-(0\,m)^{2}]}$

$v_{2}\approx 5.272\,\frac{m}{s}$

The muzzle velocity of the ball is approximately 5.272 meters per second.

3) After leaving the toy gun, the ball is solely decelerated by gravity. We construct this model by Principle of Energy Conservation:

$U_{g,2}+K_{2} = U_{g,3}+K_{3}$ (Eq. 3)

Where:

$U_{g,2}$ , $U_{g,3}$ - Initial and gravitational potential energies of the ball, measured in joules.

$K_{2}$ , $K_{3}$ - Initial and final translational kinetic energies of the ball, measured in joules.

After applying definitions of gravitational potential and translational kinetic energies, we expand the equation above and solve the resulting for the final height:

$m\cdot g \cdot (h_{3}-h_{2}) = \frac{1}{2}\cdot m \cdot (v_{2}^{2}-v_{3}^{2})$

$h_{3}-h_{2}=\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$

$h_{3} = h_{2} +\frac{v_{2}^{2}-v_{3}^{2}}{2\cdot g}$ (Eq. 4)

$h_{2}$ , $h_{3}$ - Initial and final heights of the ball, measured in meters.

$v_{2}$ , $v_{3}$ - Initial and final velocities of the ball, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

If we get that $v_{2} = 5.272\,\frac{m}{s}$ , $v_{3} = 0\,\frac{m}{s}$ , $h_{2} = 0\,m$ and $g = 9.807\,\frac{m}{s^{2}}$ , the maximum height of the ball is:

$h_{3} = 0\,m+\frac{\left(5.272\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$h_{3} = 1.417\,m$

The maximum height of the ball is 1.417 meters.

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Answer:

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