Answer:
0.752 m/s
Explanation:
m1 = 3.00kg
u1 = 5.05m/s
m2 = 2.76kg
u2 = -3.66m/s
According to the law of conservation of momentum,
m1u1 + m2u2 = (m1+m2)v
3(5.05) + 2.76(-3.66) = (5.05+2.76)v
15.15 - 9.2736 = 7.81v
5.8764 = 7.81v
v = 5.8764/7.81
v = 0.752m/s
Answer:
Speed = 575 m/s
Mechanical energy is conserved in electrostatic, magnetic and gravitational forces.
Explanation:
Given :
Potential difference, U = ![$-3.45 \times 10^{-3} \ V$](https://tex.z-dn.net/?f=%24-3.45%20%5Ctimes%2010%5E%7B-3%7D%20%5C%20V%24)
Mass of the alpha particle, ![$m_{\alpha} = 6.68 \times 10^{-27} \ kg$](https://tex.z-dn.net/?f=%24m_%7B%5Calpha%7D%20%3D%206.68%20%5Ctimes%2010%5E%7B-27%7D%20%5C%20kg%24)
Charge of the alpha particle is, ![$q_{\alpha} = 3.20 \times 10^{-19} \ C$](https://tex.z-dn.net/?f=%24q_%7B%5Calpha%7D%20%3D%203.20%20%5Ctimes%2010%5E%7B-19%7D%20%5C%20C%24)
So the potential difference for the alpha particle when it is accelerated through the potential difference is
![$U=\Delta Vq_{\alpha}$](https://tex.z-dn.net/?f=%24U%3D%5CDelta%20Vq_%7B%5Calpha%7D%24)
And the kinetic energy gained by the alpha particle is
![$K.E. =\frac{1}{2}m_{\alpha}v_{\alpha}^2 $](https://tex.z-dn.net/?f=%24K.E.%20%3D%5Cfrac%7B1%7D%7B2%7Dm_%7B%5Calpha%7Dv_%7B%5Calpha%7D%5E2%20%24)
From the law of conservation of energy, we get
![$K.E. = U$](https://tex.z-dn.net/?f=%24K.E.%20%3D%20U%24)
![$\frac{1}{2}m_{\alpha}v_{\alpha}^2 = \Delta V q_{\alpha}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B1%7D%7B2%7Dm_%7B%5Calpha%7Dv_%7B%5Calpha%7D%5E2%20%3D%20%5CDelta%20V%20q_%7B%5Calpha%7D%24)
![$v_{\alpha} = \sqrt{\frac{2 \Delta V q_{\alpha}}{m_{\alpha}}}$](https://tex.z-dn.net/?f=%24v_%7B%5Calpha%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%20%5CDelta%20V%20q_%7B%5Calpha%7D%7D%7Bm_%7B%5Calpha%7D%7D%7D%24)
![$v_{\alpha} = \sqrt{\frac{2(3.45 \times 10^{-3 })(3.2 \times 10^{-19})}{6.68 \times 10^{-27}}}$](https://tex.z-dn.net/?f=%24v_%7B%5Calpha%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B2%283.45%20%5Ctimes%2010%5E%7B-3%20%7D%29%283.2%20%5Ctimes%2010%5E%7B-19%7D%29%7D%7B6.68%20%5Ctimes%2010%5E%7B-27%7D%7D%7D%24)
![$v_{\alpha} \approx 575 \ m/s$](https://tex.z-dn.net/?f=%24v_%7B%5Calpha%7D%20%5Capprox%20575%20%5C%20m%2Fs%24)
The mechanical energy is conserved in the presence of the following conservative forces :
-- electrostatic forces
-- magnetic forces
-- gravitational forces
Answer:
![P(3600)=593.247W](https://tex.z-dn.net/?f=P%283600%29%3D593.247W)
Explanation:
First, let's find the voltage through the resistor using ohm's law:
![V=IR=20*8=160V](https://tex.z-dn.net/?f=V%3DIR%3D20%2A8%3D160V)
AC power as function of time can be calculated as:
(1)
Where:
![\phi=Phase\hspace{3}angle\\\omega= Angular\hspace{3}frequency](https://tex.z-dn.net/?f=%5Cphi%3DPhase%5Chspace%7B3%7Dangle%5C%5C%5Comega%3D%20Angular%5Chspace%7B3%7Dfrequency)
Because of the problem doesn't give us additional information, let's assume:
![\phi=0\\\omega=2 \pi f=2*\pi *(60)=120\pi](https://tex.z-dn.net/?f=%5Cphi%3D0%5C%5C%5Comega%3D2%20%5Cpi%20f%3D2%2A%5Cpi%20%2A%2860%29%3D120%5Cpi)
Evaluating the equation (1) in t=3600 (Because 1h equal to 3600s):
![P(3600)=160*8*cos(0)-160*8*cos(2*120\pi*3600-0)\\P(3600)=1280-1280*cos(2714336.053)\\P(3600)=1280-1280*0.5365255751\\P(3600)=1280-686.7527361=593.2472639\approx=593.247W](https://tex.z-dn.net/?f=P%283600%29%3D160%2A8%2Acos%280%29-160%2A8%2Acos%282%2A120%5Cpi%2A3600-0%29%5C%5CP%283600%29%3D1280-1280%2Acos%282714336.053%29%5C%5CP%283600%29%3D1280-1280%2A0.5365255751%5C%5CP%283600%29%3D1280-686.7527361%3D593.2472639%5Capprox%3D593.247W)
Answer: 113.75
Explanation:
You know
acceleration = a = 3.5 m/s²
time = t = 5 seconds
initial velocity = u = 14 m/s
Unknown is distance = s = ?
Use equation: s = ut +
at²
Substitute all the known values inside the equation:
s = (14*5) + 0.5 * 3.5 * 5²
s = 70 + 43.75 = 113.75 m
The car travels 113.75 metres.