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ICE Princess25 [194]

2 years ago

5

When a quarterback throws a football, the throwing arm is slowed down to rest during the deceleration phase. Of the angular velo

city of the throwing arm is 1800 deg/sec at the instant of ball release, then how much would the arm deceleration change if the deceleration was reduced from 2 sec to 0.5 sec

1 answer:

asambeis [7]2 years ago

3 0

Answer:

The deceleration is 4 times higher for 0.5sec than for 2 sec.

Explanation:

If the deceleration happens from 1800deg/sec to zero in 2 sec we have that the magnitude is:

$a_{1}=w/t_{1}=(1800deg/sec)/2sec=900 deg/sec^{2}$

if now the time to decelerate is 0.5sec:

$a_{2}=w/t_{2}=(1800deg/sec)/0.5sec=3600 deg/sec^{2}$

The deceleration is now 4 times higher than the previous situation.

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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

pishuonlain [190]

Answer:

$E_{total}=4.82*10^6N/C$

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

<u>Superposition Law:</u> the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

$E_{total}=E_1+E_2$

<u>Thanks Gauss Law</u> we know that the electric field of a infinite sheet with density of charge σ is:

$E=\sigma/(2\epsilon_o)$

Then:

$E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C$

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0

2 years ago

Read 2 more answers

What did you notice about the angle of incidence and the angle of reflection?

Readme [11.4K]

Answer:

They are equal

Explanation:

angle of incidence = angle of reflection

6 0

1 year ago

A penny rides on top of a piston as it undergoes vertical simple harmonic motion with an amplitude of 4.0cm. If the frequency is

aniked [119]

1) At the moment of being at the top, the piston will not only tend to push the penny up but will also descend at a faster rate at which the penny can reach in 'free fall', in that short distance. Therefore, at the highest point, the penny will lose contact with the piston. Therefore the correct answer is C.

2) To solve this problem we will apply the equations related to the simple harmonic movement, hence we have that the acceleration can be defined as

$a = -\omega^2 A$

Where,

a = Acceleration

A = Amplitude

$\omega$ = Angular velocity

From a reference system in which the downward acceleration is negative due to the force of gravity we will have to

$a = -g$

$-\omega^2 A = -g$

$\omega = \sqrt{\frac{g}{A}}$

From the definition of frequency and angular velocity we have to

$\omega = 2\pi f$

$f = \frac{1}{2\pi} \sqrt{\frac{g}{A}}$

$f = \frac{1}{2\pi} \sqrt{\frac{9.8}{4*10^{-2}}}$

$f = 2.5Hz$

Therefore the maximum frequency for which the penny just barely remains in place for the full cycle is 2.5Hz

6 0

2 years ago

What is the wavelength, in nm, of the line in the hydrogen spectrum when one n value is 3 and the other n value is 6?

iren [92.7K]

Answer:

$\lambda=1090nm$

Explanation:

Rydberg formula is used to calculate the wavelengths of the spectral lines of many chemical elements. For the hydrogen, is defined as:

$\frac{1}{\lambda}=R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

Where $R_H$ is the Rydberg constant for hydrogen and $n_1$ , $n_2$ are the lower energy state and the higher energy state, respectively.

$\frac{1}{\lambda}=1.10*10^{7}m^{-1}(\frac{1}{3^2}-\frac{1}{6^2})\\\frac{1}{\lambda}=9.17*10^{5}m^{-1}\\\lambda=\frac{1}{1.09*10^{6}m^{-1}}\\\lambda=1.09*10^{-6}m*\frac{10^{9}nm}{1m}\\\lambda=1090nm$

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3 years ago

Questions for 1.21 physics lab report

Lapatulllka [165]

Ok cool dude bro I just need to answer a question

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3 years ago