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2 years ago

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A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 me

ters per second. What is the magnitude of the total momentum of the two balls after collision?

1 answer:

enyata [817]2 years ago

6 0

Answer:

We know the momentum after the collision MUST be equal to the momentum BEFORE the collision.

Momentum is a VECTOR quantity having both magnitude and direction. The first ball has momentum P =m*v = 2*4 = 8 at 90degrees. The second ball has momentum P = 1*8 = 8 at -90 or 270 degrees. They sum to zero when you perform vector addition.

Explanation:

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A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o

Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field $dE$ at the center of curvature of the arc is

$dE = k\dfrac{dQ}{r^2}cos(\theta )$ <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where $r$ is the radius of curvature.

Now

$dQ = \lambda rd\theta$ ,

where $\lambda$ is the charge per unit length, and it has the value

$\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.$

Thus, the electric field at the center of the curvature of the arc is:

$E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta }{r^2} cos(\theta)$

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.$

Now, we find $\theta_1$ and $\theta_2$ . To do this we ask ourselves what fraction is the arc length 3.0 of the circumference of the circle:

$fraction = \dfrac{3.0m}{2\pi (2.3m)} = 0.2076$

and this is

$0.2076*2\pi =1.304$ radians.

Therefore,

$E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.$

evaluating the integral, and putting in the numerical values we get:

$E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\$

$\boxed{ E = 31.329N/C.}$

4 0

3 years ago

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas

marshall27 [118]

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1+P_2V_2=P_fV_f$

where,

$P_1$ = first pressure = 8.19 atm

$P_2$ = second pressure = 2.65 atm

$V_1$ = first volume = 2.14 L

$V_2$ = second volume = 9.84 L

$P_f$ = final pressure = ?

$V_f$ = final volume = 2.14 L + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

$8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L$

$P_f=3.64atm$

Therefore, the final pressure of the system in atm is, 3.64 atm

4 0

3 years ago

Cocking your head would be most useful for detecting the ______ of a sound.

statuscvo [17]

Cocking your head would be most useful for detecting the LOCATION of a sound.

5 0

3 years ago

Read 2 more answers

Why is Carbon Tetrahydride a covalent bond? plzzz helpppp!!!!!!

arlik [135]

Answer:

carbon has four unpaired electrons in its valence shell . hydrogen having one unpaired electron in its valence shell comes to bond with carbon by sharing a pair of electrons .since carbon needs 4 electrons to be stable, 4 hydrogen atoms take part in the bond . It's a covalent bond because the difference between the electronegativity of carbon and hydrogen is quite small .

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2 years ago

Elodia [21]

Answer: 15.29

Explanation: there you go have a nice day (*^p^*)

3 0

2 years ago