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3 years ago

Como se conserva la energia mecanica del cuerpo de un deportista

Physics

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

the energy food has is transformed into mechanical energy during exercise,

Explanation:

When a person is exercising, the energy accumulated in food is transformed into mechanical energy, in general if he is running in kinetic energy and if he is rising and falling in potential energy.

In this case the energy food has is transformed into mechanical energy during exercise,

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Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp

slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

$Q=Av$

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

$Q=1.20 m^3/s$

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

$A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2$

So we can re-arrange the equation to find the speed of the water:

$v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s$

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

$Q_1 = Q_2\\A_1 v_1 = A_2 v_2$

where we have

$A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s$

and where $A_2$ is the cross-sectional area of the pipe at the second point.

Solving for A2,

$A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2$

And finally we can find the radius of the pipe at that point:

$A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m$

6 0

3 years ago

The type of force measured by grocery store spring scale is

alexandr402 [8]

Answer:

Weight

Explanation:

The spring balance is used to measure weight of an object.

8 0

2 years ago

A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f

sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

$F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)$

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

5 0

2 years ago

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

igomit [66]

Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is $mgsin \alpha$ . Therefore the effort force in this case must be equal or greater than $mgsin \alpha$ .

Now we need to find $\alpha$ . $\alpha$ is the angle between the incline of the ramp and the ground.

Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

=> m*g* $sin \alpha$ = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
$F_{e} d_{e} = F_{r} d_{r}$

$F_{e}$ = ?

$F_{r}$ = mg = 20 * 10 = 200N
$d_{e}$ = 10m
$d_{r}$ = 1m

Plug-in the values in the above equation:
$F_{e}$ = 200/10= 20N

As 20N << 125N, the best choice is to use lever.

4 0

3 years ago

Jacob is traveling at 5.00 m/s North. Jacob throws a ball with a velocity of 5.00 m/s South. Jacob throws the ball from a height

Nataly [62]

Answer:t=0.54 s

Explanation:

Given

Jacob is traveling 5 m/s in North direction

Jacob throw a ball with a in south direction with a velocity of 5 m/s

Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0

Time taken by ball to reach ground

$s=ut+\frac{gt^2}{2}$