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zzz [600]
3 years ago
7

an elevator mass of 7700 kg falls from a height of 32 m after a sudden failure in the hoisting cable. The mass is stopped by a s

pring at the bottom of the shaft. Determine the spring constant (in kN/m) necessary to bring the elevator and occupants to rest without exceeding an acceleration of 5 g's.
Physics
1 answer:
valkas [14]3 years ago
4 0

Answer:k=28.29 kN/m

Explanation:

Given

mass m =7700 kg

height from which Elevator falls h=32 m

Let x be the compression in the spring

thus From conservation of Energy Potential energy will convert in to Elastic Potential Energy of spring

\frac{kx^2}{2}=mg(h+x)----------1

also maximum acceleration is 5g

thus

mg-kx=ma

here a=-5g

kx=mg-m(-5g)=6mg

x=\frac{6mg}{k}

Substitute x in equation 1

0.5\times k\times (\frac{6mg}{k})^2=mg(h+\frac{6mg}{k})

18\frac{(mg)^2}{k}=mgh+6\frac{(mg)^2}{k}

k=12\cdot \frac{mg}{h}

k=12\times \frac{7700\times 9.8}{32}

k=28.29 kN/m

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Two lightweight rods 24 cm in length are mounted perpendicular to an axle and at 180 ∘ to each other. At the end of each rod is
nikitadnepr [17]

Answer:

Explanation:

Angular momentum ( L ) = moment of inertia x angular velocity ( I X ω )

Moment of inertia of two  480 g masses about axle = 2 x mr²  = 2 x 480 x10⁻³ x( 24 x 10 ⁻ 2 )²  = 0. 552960  kg m².

Angular velocity = 5 rad / s.

Angular momentum = 0.552960 x 5 = 2.765 kg m2.

The direction of angular momentum will be along axle.So vector angular

momentum  makes zero degree  with axle.

7 0
3 years ago
How do H-R diagrams help scientists compare new stars with the Sun? In your answer, name one type of data that H-R diagrams do n
damaskus [11]

Answer

Hertzsprung-Russell (HR) diagram is an essential tool used in stellar evolution. In the universe, there are several hundreds of billions of stars. Scientists use the tool, in differentiation, the billions of stars in the world from the sun. In the HR tool, there is plotting of the luminosity or energy output of a star, which is plotted on the X-axis of a graph against the absolute magnitude. The sun's magnitude is an absolute of +48, which, when plotted against its luminosity, helps in setting an apparent variance between the sun and any other star. Additionally, the sun has been identified as the primary star with a very high temperature. Hence the tool can locate the sun from other forms of stars. HR diagrams outline data such as temperature and luminosity or energy. However, star distance from the Erath is not a type of data represented in the charts.

Explanation:

Hope this helped you!

4 0
3 years ago
Read 2 more answers
Find the quantity of heat needed
krok68 [10]

Answer:

Approximately 3.99\times 10^{4}\; \rm J (assuming that the melting point of ice is 0\; \rm ^\circ C.)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}

The energy required comes in three parts:

  • Energy required to raise the temperature of that 0.100\; \rm kg of ice from (-10\; \rm ^\circ C) to 0\; \rm ^\circ C (the melting point of ice.)
  • Energy required to turn 0.100\; \rm kg of ice into water while temperature stayed constant.
  • Energy required to raise the temperature of that newly-formed 0.100\; \rm kg of water from 0\; \rm ^\circ C to 10\;\ rm ^\circ C.

The following equation gives the amount of energy Q required to raise the temperature of a sample of mass m and specific heat capacity c by \Delta T:

Q = c \cdot m \cdot \Delta T,

where

  • c is the specific heat capacity of the material,
  • m is the mass of the sample, and
  • \Delta T is the change in the temperature of this sample.

For the first part of energy input, c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}.

Similarly, for the third part of energy input, c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1} whereas m = 0.100\; \rm kg. Calculate the change in the temperature:

\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}.

Calculate the energy required to achieve that temperature change:

\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}.

The second part of energy input requires a different equation. The energy Q required to melt a sample of mass m and latent heat of fusion L_\text{f} is:

Q = m \cdot L_\text{f}.

Apply this equation to find the size of the second part of energy input:

\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}.

Find the sum of these three parts of energy:

\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}.

3 0
2 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
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This is the first stanza of the song:

"I was a highwayman. Along the coach roads I did ride 
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<span>Many a soldier shed his lifeblood on my blade </span>
<span>The b*stards hung me in the spring of twenty-five </span>
<span>But I am still alive."</span>
 
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