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3 years ago

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Would You Rather...Carry a 50 lb box up 3 flights of stairs or a 25 lb box up 5 flights of stairs? Whichever choice you make, ju

stify your reasoning with mathematics and/or physics responses. *

1 answer:

Oksanka [162]3 years ago

8 0

I would carry a 25 pound box up 5 flights, because it is lighter. You will wear out faster carrying a 50 pound box up 3 flights.

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Spinning situations???

Papessa [141]

Just like mass, energy, linear momentum, and electric charge, angular momentum is also conserved.

The wheel has angular momentum. I don't remember whether it's
up or down (right-hand or left-hand rule), but it's consistent with
counterclockwise rotation as viewed from above.

When you grab the wheel and stop it from spinning (relative to you),
that angular momentum has to go somewhere.

As I see it, the angular momentum transfers through you as a temporary
axis of rotation, and eventually to the merry-go-round. Finally, all the mass
of (merry-go-round) + (you) + (wheel) is rotating around the big common
axis, counterclockwise as viewed from above, and with the magnitude
that was originally all concentrated in the wheel.

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3 years ago

Read 2 more answers

A motorcycle has a mass of 2.50×10^2. A constant force is exerted on it for 60.0s. The motorcycles initial velocity is 6.00 m/s

Margarita [4]

The change in momentum is 5500 kg m/s

Explanation:

The change in momentum of an object is given by

$\Delta p = m(v-u)$

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have:

$m=2.50\cdot 10^2 kg$ (mass of the motorcycle)

$v=28.0 m/s$ (final velocity)

$u=6.0 m/s$ (initial velocity)

Therefore, the change in momentum is

$\Delta p=(2.50\cdot 10^2)(28.0-6.0)=5500 kg m/s$

Learn more about change in momentum:

brainly.com/question/9484203

#LearnwithBrainly

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3 years ago

Explain Rutherford's experiment?

Ipatiy [6.2K]

Answer:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

Explanation:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.

The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.

Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.

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3 years ago

What happens to jetstream’s as they get closer to the equator

MAVERICK [17]

Answer:They stop because jet streams follow boundaries between hot and cold air.

Explanation:

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3 years ago

Read 2 more answers

A spring gun is made by compressing a spring in a tube and then latching the spring at the compressed position. A 4.97-g pellet

dimaraw [331]

Answer:

$v = 2.8898 \frac{m}{s}$

Explanation:

This is a problem easily solve using energy conservation. As there are no non-conservative forces, we know that the energy is conserved.

When the spring is compressed downward, the spring has elastic potential energy. When the spring is relaxed, there is no elastic potential energy, but the pellet will have gained gravitational potential energy and kinetic energy. Lets see what are the terms for each of this.

<h3>Elastic potential energy</h3>

We know that a spring following Hooke's Law has a elastic potential energy:

$E_{ep} = \frac{1}{2} k (\Delta x)^2$

where $\Delta x$ is the displacement from the relaxed length and k is the spring's constant.

To obtain the spring's constant, we know that Hooke's law states that the force made by the spring is :

$\vec{F} = - k \Delta \vec{x}$

as we need 9.12 N to compress 4.60 cm, this means:

$k = \frac{9.12 \ N}{4.6 \ 10^{-2} \ m}$

$k = 198.26 \ \frac{ N}{m}$

So, the elastic energy of the compressed spring is:

$E_{ep} = \frac{1}{2} 198.26 \ \frac{ N}{m} (4.6 \ 10^{-2} \ m)^2$

$E_{ep} = 0.209759 \ Joules$

And when the spring is relaxed, the elastic potential energy will be zero.

<h3>Gravitational potential energy</h3>

To see how much gravitational potential energy will the pellet win, we can use

$\Delta E_{gp} = m g \Delta h$

where m is the mass of the pellet, g is the acceleration due to gravity and \Delta h is the difference in height.

Taking all this together, the gravitational potential energy when the spring is relaxed will be:

$\Delta E_{gp} = 4.97 \ 10^{-3} kg \ 9.8 \frac{m}{s^2} 4.6 \ 10^{-2} m$

$\Delta E_{gp} = 0.00224 \ Joules$

<h3>Kinetic Energy</h3>

We know that the kinetic energy for a mass m moving at speed v is:

$E_k = \frac{1}{2} m v^2$

so, for the pellet will be

$E_k = \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

<h3>All together</h3>

By conservation of energy, we know:

$E_{ep} = \Delta E_{gp} + E_k$

$0.209759 \ Joules = 0.00224 \ Joules + \frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2$

So

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.209759 \ Joules - 0.00224 \ Joules$

$\frac{1}{2} \ 4.97 \ 10^{-3} kg \ v^2 = 0.207519 \ Joules$

$v = \sqrt{ \frac{ 0.207519 \ Joules}{ \frac{1}{2} \ 4.97 \ 10^{-3} kg } }$

$v = 2.8898 \frac{m}{s}$

7 0

3 years ago