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4 years ago

A student weighing 120 lbs climbs a 12 ft flight of stairs in 9 seconds. how much power did the student create?

1 answer:

7 0

Power can be calculate through the equation,
Power = Force x velocity

It should be noted that velocity is calculated by dividing displacement by time. Thus, from the given in this item we can calculate for the power.
Power = (120 lb) x (12 ft/9 s)
 Power = 160 lb.ft/s

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svetoff [14.1K]

D.) In order to calculate both of them, we must know the "FORCE" on the system.

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3 years ago

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If you ride your bike at an average of 7km/h and need to travel a total distance of 42km how long will it take you to reach your

alekssr [168]

Answer:

42÷7

Explanation:

6 hours

since 7km is 1hour then

42km will be 6 hours

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3 years ago

What items can be classified as matter?

Agata [3.3K]

Gas , Liquid , or solid

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4 years ago

A (hypothetical) large slingshot is stretched 4.00 m to launch a 440 g projectile with speed sufficient to escape from Earth (11

RSB [31]

Answer:

(A) 3,449,600 N/m

(B) 62,720 poeple

Explanation:

extension of the sling shot (e) = 4 m

mass of projectile (m) = 440 g = 0.44 kg

speed of projectile (v) = 11.2 km/s = 11,200 m/s

average force a person can exert = 220 N

spring constant (k) = ?

(A) When all the elastic potential energy is converted to kinetic energy

$o.5ke^{2} = 0.5 mv^{2}$

rearranging the above equation

spring constant (K) = $\frac{0.5mv^{2} }{0.5e^{2}}$

spring constant (K) = $\frac{0.5x0.44x11,200^{2} }{0.5x4^{2}}$

spring constant (K) = $\frac{27,596,800}{8}$

spring constant (K) = 3,449,600 N/m

(B) force required to stretch the slingshot (F) = ke = 3,449,600 x 4 = 13,798,400 N

number of people required = required force / average force per person = 13,798,400 / 220 =62,720 poeple

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3 years ago

Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum

Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

$\frac{1}{2}mv^2 = mgh$

where

m is the mass of the man

v is the speed after jumping

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

$v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s$

In the acceleration phase, we know that the initial velocity is

$u=0$

And the force exerted on the floor is 2.3 times the gravitational force, so

$F=2.3 mg$

This means the net force on you is

$F_{net} = F-mg=2.3mg-mg=1.3 mg$

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

$a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g$

Now we can use the following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

$v^2-u^2=2as$

where s is the quantity we want to find. Solving for s,

$s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m$

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

$\Delta h = -0.77 m$

This means that the lowest point reached by the center of mass above the floor during the crouch is

$h'=h+\Delta h = 1.0 - 0.77 = 0.23 m$

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0

3 years ago