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garik1379 [7]
3 years ago
11

Chan Hee is inflating a basketball using an air pump. He notices that the pump gets warm as he uses it. What is a good hypothesi

s that could lead to new experimentation?
If there is no air in a basketball, then it is hard to play the game.
A basketball that is inflated bounces better than one that is not inflated.
If air in a pump is squeezed more, then the air gets hotter because energy is added to it.
Basketball is more fun to play than baseball because you have to move a lot more.
Physics
1 answer:
likoan [24]3 years ago
8 0

"If air in a pump is squeezed more, then the air gets hotter because energy is added to it" is a good hypothesis that could lead to new experimentation.

<u>Option: C</u>

<u>Explanation:</u>

If we use a pump to inflate a basketball, we initially pull the handle to draw air to fill the sphere in. As we move it down we apply a great deal of force to pump in air through the pin's tiny hole because of this resistance force in the air we find the tube warmed.

A needle of ball pump is a metal tube in which air, from an inflating pump to a sports ball, moves through it. In continuous-flow operation, pumps are often used and built to produce comparatively little pressure towards a free-flowing environment with limited back pressure. Such pumps have a fixed configuration and work freely along their power curve as circumstances change.

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A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment
snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = I\alpha

\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2

Now

\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians

Here t = 1 minutes = 60 seconds

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3 years ago
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4 0
3 years ago
Read 2 more answers
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
4 years ago
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Anastasy [175]

Answer:

omg this is hard

Explanation:

Yes it is lol

5 0
3 years ago
Melting Ice is what kind of reaction
Vesnalui [34]

Answer:

Answer is Endothermic Reaction

Explanation:

Basically, melting ice is an endothermic reaction because the ice absorbs (heat) energy, which causes a change to occur.

I hope it's helpful!!

7 0
3 years ago
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