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2 years ago

You carry a 20 n bag of dog food up a 6 m flight of stairs. how much work was done?

Physics

1 answer:

ryzh [129]2 years ago

5 0

Answer:

No work was done because it is not hard to carry a ####### bag of dog food up the stairs.

Explanation:

The FitnessGram Pacer Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The 20 meter pacer test will begin in 30 seconds. Line up at the start. The running speed starts slowly, but gets faster each minute after you hear this signal. A single lap should be completed each time you hear this sound. Remember to run in a straight line, and run as long as possible.

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A force of 70 N is applied to a 28 kg rock causing it to slow down from 25 m/s to 15 m/s, a change in velocity of 10 m/s. How lo

maksim [4K]

Answer: 4 s

Explanation:

Given

The applied force is 70 N

mass of the rock is 28 kg

initial velocity $u=25\ m/s$

final velocity $v=15\ m/s$

Deceleration provided by force is

$a=-\dfrac{70}{28}=-2.5\ m/s^2$

using the equation of motion

$v=u+at\\\Rightarrow 15=25-2.5t\\\Rightarrow 2.5t=10\\\Rightarrow t=4\ s$

7 0

3 years ago

A person, with a mass of 50.0 kg, stands on a weighing scale in a lift which is moving

svet-max [94.6K]

Answer:

Explanation:

Because I just had that answer

6 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .