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3 years ago

How come walking converts chemical energy into mechanical energy

2 answers:

6 0

Walking requires mechanical energy, because it involved accelerating and lifting a mass. That energy has to come from somewhere. Your body isn't built to run on solar panels, or a windmill, or a nuclear reactor, so the only place you can get the energy is from putting food into yourself and letting your stomach and guts do chemical reactions with it.

tekilochka [14]3 years ago

3 0

Chemical energy (calories) is converted by your body walking on the surface into mechanical/kinetic energy

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How would you respond to roger?

ElenaW [278]

Hello Roger. How you be?

8 0

3 years ago

A m = 500 kg sack of coal falls vertically onto a m = 2000 kg railroad flatcar which was initially moving horizontally at 3 m/s.

vladimir2022 [97]

You can view this as an inelastic collision between a 2000 kg object moving at 3 m/s, and a stationary 500 kg object.
The total momentum of the objects before the collision equals their momentum after collision.
Before:
flatcar = 2000*3 = 6000 kg*m/s.
<span> coal = 0*500 = 0.
After:
flatcar & coal = 2500*V
then
</span> 2500*V <span>= 6000,
V = 2.4 m/s.
</span> Additional:
the definition of an inelastic collision is that the objects "stick together" after colliding.

4 0

4 years ago

Read 2 more answers

Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same

Contact [7]

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

$\underset{V_{A}}{\rightarrow}$ = velocity of car A before collision = 0 i - $V_{A}$ j

$\underset{V_{B}}{\rightarrow}$ = velocity of car B before collision = $V_{B}$ i + 0 j

$\underset{V_{AB}}{\rightarrow}$ = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

$M_{A}$ = mass of car A = 1750 kg

$M_{B}$ = mass of car B = 1450 kg

Using conservation of momentum

$M_{A}$ $\underset{V_{A}}{\rightarrow}$ + $M_{B}$ $\underset{V_{B}}{\rightarrow}$ = ( $M_{A}$ + $M_{B}$ ) ( $\underset{V_{AB}}{\rightarrow}$ )

(1750) (0 i - $V_{A}$ j) + (1450) ( $V_{B}$ i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) $V_{B}$ i - (1750) $V_{A}$ j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) $V_{B}$ = 97600 and - (1750) $V_{A}$ = - 60160

$V_{B}$ = 67.3 km/h and $V_{A}$ = 34.4 km/

6 0

4 years ago

In a softball game, a batter hits the ball at the velocity and angle shown

Kipish [7]

This problem is going to be pretty long to solve. So, prepare.

We’re interested in the change in our x position. So we have to break the velocity vector up into its components. Do cosine of 50 and then multiply by the magnitude of the velocity. I got 20.57m/s. That’s our initial velocity. And remember, horizontal acceleration is zero. The vertical acceleration, or any vertical component, has no effect on the horizontal components. In order to solve this problem, we want to utilize this equation:
Change in x-position = Vix*t

Let’s solve for time, which is dependent on the vertical components. The projectile will stop when it vertically hits the ground. Generally you want to use this equation for solving for time:
Yf = Yi + Viy*t + 1/2at^2
We didn’t solve for the vertical component yet, so let’s do that now. (Sine of 50)*(32) = 24.51m/s
Let’s now plug everything in:
0 = 0 + 24.5t - 4.9t^2
0 = 24.5t - 4.9t^2
0 = t(24.5 - 4.9t)
-24.5 = -4.9t
t = 5 seconds

The hard stuff is pretty much over. Put that 5 seconds into the other equation I said we wanted to use to solve the problem
Change in x-position (range) = (20.57)*(5)
= 102.85 meters

Answer B

5 0

4 years ago

. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin

Viktor [21]

Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

$T-(m*g)=m*a$

Where:

$T=$ thrust $=3 *10^{4} N$

$m=$ module's mass $=1 *10^{4} N$

$g=$ moon's gravity acceleration $=1.623 \frac{m}{s^2}$

$a=$ module's acceleration during takeoff

Then, we can find the acceleration like this:

$a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}$

$a=1.377 \frac{m}{s^2}$

The module's acceleration in a vertical takeoff from the Moon will be $1.377 \frac{m}{s^2}$

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

$weight=m*g=(1*{10}^4 kg)*(9.8 \frac{m}{s^2})=9.8 *10^{4} N$

Then we can say that a thrust of $3*10^{4} N$ won't be able to lift off the module from the Earth because it's smaller than the module's weight ( $9.8 *10^{4} N$ ).

8 0

4 years ago