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4 years ago

Real world situation to explain the relationship between work and power

1 answer:

5 0

-- You and your partner both get the same job to do:

Each of you gets a pallet of bricks, and you have to
put the bricks up on the bed of a truck, by hand.
Both pallets have the same number of bricks.

The pallet is way too heavy to lift, so you both cut the bands
that hold the bricks, and you lift the bricks from the pallet onto
the truck, by hand, two or three or four bricks at a time.

-- You get your pallet of bricks onto the truck in 45 minutes.

-- Your partner gets his pallet of bricks onto the truck in 3 days.

-- Work = (force) times (distance).

    You and your partner both lifted the same amount of weight
   up to the same height. You both did the same amount of work.

-- Power = (work done) divided by (time it takes to do the work) .

   Your partner took roughly 96 times as long as you took
   to do the same amount of work.
   You did it faster. He did it slower.
   You produced more power. He produced less power.

You might be interested in

A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 g

alekssr [168]

Answer:

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

--------------------------------------------------------------------------

↓ Δ ↓

20 g.................50 cm.................40g

38 cm y cm

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

12cm 50 cm 69cm

--------------------------------------------------------------------------

↓ Δ ↓

20 g.................50 cm.................40g

38 cm 19 cm

5 0

3 years ago

What is net force?

kramer

Answer:

A. The sum of all the forces acting on an object.

3 0

3 years ago

What are the forces that act on the ball?

scoundrel [369]

Answer:

This slide shows the three forces that act on a baseball in flight. The forces are the weight, drag, and lift. Lift and drag are actually two components of a single aerodynamic force acting on the ball. Drag acts in a direction opposite to the motion, and lift acts perpendicular to the motion

4 0

3 years ago

A projectile is fired from a height of 80 M above sea level, horizontally with a speed of 360 M / S, calculate: The time it take

Maslowich

Answer:

(a) The projectile takes approximately 4.420 seconds to reach the water, (b) The horizontal scope of the projectile is 1591.2 meters, (c) The remaining height to descend after 2 seconds of being launched is 63.624 meters.

Explanation:

The projectile experiments a parabolic motion, where horizontal speed remains constant and accelerates vertically due to the gravity effect. Let consider that drag can be neglected, so that kinematic equation are described below:

$x = x_{o}+v_{o,x} \cdot t$

$y = y_{o} + v_{o,y}\cdot t +\frac{1}{2}\cdot g \cdot t^{2}$

Where:

$x_{o}$ , $y_{o}$ - Initial horizontal and vertical position of the projectile, measured in meters.

$v_{o,x}$ , $v_{o,y}$ - Initial horizontal and vertical speed of the projectile, measured in meters per second.

$t$ - Time, measured in seconds.

$g$ - Gravitational acceleration, measured in meters per square second.

$x$ , $y$ - Current horizontal and vertical position of the projectile, measured in meters.

Given that $x_{o} = 0\,m$ , $y_{o} = 80\,m$ , $v_{o,x} = 360\,\frac{m}{s}$ , $v_{o,y} = 0\,\frac{m}{s}$ and $g = -9.807\,\frac{m}{s^{2}}$ , the kinematic equations are, respectively:

$x = 360\cdot t$

$y = 80-4.094\cdot t^{2}$

(a) If $y = 0\,m$ , the time taken for the projectile to reach the water is:

$80 - 4.094\cdot t^{2} = 0$

$t = \sqrt{\frac{80}{4.094} }\,s$

$t \approx 4.420\,s$

The projectile takes approximately 4.420 seconds to reach the water.

(b) The horizontal scope is the horizontal distance done by the projectile before reaching the water. If $t \approx 4.420\,s$ , the horizontal scope of the projectile is:

$x = 360\cdot (4.420)$