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3 years ago

It is easier to climb up a slanted slope than a vertical slope

2 answers:

8 0

Yes it is easier to clime a gentle slope than a vertical or more steep slope. This is because gravity is always pulling objects down so when you climb upward gravity weighs you down, you have to use effort to move yourself in the opposite direction that gravity is pulling you that is why it is harder to climb up a slope. When you climb a gentle slope you still are using energy but it takes longer to reach the same height on a gentle slope versus a vertical slope. On a vertical slope you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope. When climbing a gentle slope you don't rise as rapidly so gravity doesn't take as much of a toll on you and you have a chance to recover.

Bas_tet [7]3 years ago

7 0

Answer:

Yes, it is easier to climb a slanted slope than a vertical or more steep slope.

Explanation:

On a vertical slope, you are climbing higher instead of farther so on each step gravity weighs you down much more than on a gentle slope

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2 years ago

Imagine that while you and a passenger are in a deep-diving submersible in the North Pacific near Alaska’s Aleutian Islands, you

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I would say its a deep ocean trench

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3 years ago

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3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$

We already know the last quantity: $\frac{m}{2} *(vel_1^2-vel_2^2)$ = $-$ Δ ( $\frac{mvel^2}{2}$ )=23400W

For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that, $h_1=h(T=500C,P=4MPa)=3445.3\frac{kJ}{kg}$

The exit state is a liquid-vapor mixture, so its enthalpy is:

$h_2=h_f+xh_{fg}=289.23+0.92*2366.1=2483.4\frac{kJ}{kg}$

Finally, the work can be obtained:

$W=12\frac{kg}{s}*(3445.3-2438.4)\frac{kJ}{kg} +23.400kW)=12106.2kW$

C) For the area, consider the equation of mass flow:

$m=p*vel*A$ where $p$ is the density, and A the area. The density is the inverse of the specific volume, so $m=\frac{vel*A}{v}$

The specific volume of the inlet steam can be read also from the steam tables, and its value is: $0.08643\frac{m^3}{kg}$ , so:

$A=\frac{m*v}{vel}=\frac{12\frac{kg}{s}*0.08643\frac{m^3}{kg}}{80\frac{m}{s}}=0.01297m^2$

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3 years ago

The circumference of a bicycle wheel is 2 meters. if it rotates at 1 revolution per second then its linear speed is

tester [92]

Given:
Circumference = 2 m
Angular speed, ω = 1 rev/s = 2π radians/s

If the radius is r, then
2πr = 2
r = 1/π m

The linear (tangential) speed is
v = rω
= (1/π m)*(2π rad/s) = 0.5 m/s

Answer: 0.5 m/s

6 0

3 years ago

It is easier to climb up a slanted slope than a vertical slope​

It is easier to climb up a slanted slope than a vertical slope