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Svetllana [295]

2 years ago

12

How can a cyclist minimize friction as he or she rides? A. by increasing the weight of the bike B. by increasing the tread on hi

s or her tires C. by decreasing the air resistance D. by decreasing aerodynamic design

2 answers:

erastova [34]2 years ago

7 0

The answer is B becausew the others would add friction

jek_recluse [69]2 years ago

6 0

B I believe is the answer!

Hope this helps and have a great day!!!

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Suppose you wanted to break a meter stick over your knee. The cross-section of a meter stick is rectangular. Will it be easier t

LenaWriter [7]

Answer:

It will be easier to break the meter rule with the long side against my knee.

Explanation:

To break the meter rule involves the principle of bending moment. The long side will require less force to generate the same amount of bending moment that will have to be generated to break the meter rule. The short side on the other hand will require more force to generate this mount of bending moment. This is because the shorter has a very small surface area, which concentrates the force on your knee. The pressure is then dissipated as more pressure to your knee. Th longer side has a lesser surface area so, most of the force is used in breaking the meter rule.

6 0

3 years ago

Which compound should require the highest temperature to melt? LiF, NaF, LiCl, NaCl

Jlenok [28]

Nacl is the answer hahahahaah ha

5 0

3 years ago

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

3 years ago

Aristotle said that a moving earthly or `mundane' object with nothing pushing or pulling on it will always A slow down and stop.

andrew11 [14]

Answer:

A slow down and stop

Explanation:

When there is no force acting on something it automatically begins to slow down and then stops.Essentially, Aristotle's perspective of motion is that "it requires a force to move an object in an unnatural" way— or, plainly, that "movement involves strength." Indeed, if you propel a book, it keeps moving. Once you stop trying to push, it comes to a stop.

5 0

2 years ago

Is physical science part of physics?

telo118 [61]

Yes, it is. Physical science<span>, the systematic study of the inorganic world</span>

6 0

3 years ago

Read 2 more answers