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3 years ago

Genetics is the most important factor that affects your physical fitness. a. True b. False

Physics

1 answer:

SIZIF [17.4K]3 years ago

6 0

It is False it's about how hard you work

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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

A uniform rod of length 0.8 m and mass 1.4 kg, has two point masses at each end. The point mass on the left end has a mass 1.2 k

VladimirAG [237]

Answer:

Explanation:

1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m

5 0

3 years ago

You throw a ball upward with an initial speed of 4.3 m/s. When it returns to your hand 0.88 s later, it has the same speed in th

djverab [1.8K]

Answer:

The acceleration is -9.8 m/s²

Explanation:

Hi there!!

When you throw a ball upward, there is a downward acceleration that makes the ball return to your hand. This acceleration is produced by gravity.

The average acceleration is calculated as the variation of the speed over time. In this case, we know the time and the initial and final speed. Then:

acceleration = final speed - initial speed/ elapsed time

acceleration = -4.3 m/s - 4.3 m/s / 0.88 s

acceleration = -9.8 m/s²

4 0

3 years ago

Which variable is not required to calculate the gibbs free-energy change?

xxMikexx [17]

All spontaneous processes release free energy

3 0

3 years ago

if vector u has lenght 70 and direction 40 degrees, and vector v has length 85 and direction 335 degrees what is the length and

Anastaziya [24]

Answer:

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

Explanation:

||u|| = 70

θ = 40°

$\vec{u}_x=||u||cos\theta \\\Rightarrow \vec{u}_x=70cos40=53.62$

$\vec{u}_y=||u||sin\theta \\\Rightarrow \vec{u}_y=70sin40=44.99$

||v|| = 85

θ = 335°

$\vec{v}_x=||v||cos\theta \\\Rightarrow \vec{v}_x=85cos335=77.03$

$\vec{v}_y=||v||sin\theta \\\Rightarrow \vec{v}_y=85sin335=-35.92$

Resultant

$R=\sqrt{R_x^2+R_y^2}\\\Rightarrow R=\sqrt{(\vec{u}_x+\vec{v}_x)^2+(\vec{u}_y+\vec{v}_y)^2}\\\Rightarrow R =\sqrt{(70cos40+85cos335)^2+(70sin40+85sin335)^2}\\\Rightarrow R =131.15$

$\theta=tan^{-1}\frac{R_y}{R_x}\\\Rightarrow \theta=tan^{-1}\frac{70sin40+85sin335}{70cos40+85cos335}\\\Rightarrow \theta=tan^{-1}0.069=3.97^{\circ}$

Magnitude of resultant = 131.15

Direction of resultant = 3.97°

4 0

3 years ago