66° N and 90° N
the area of the artic circle in the northern hemisphere
Answer:
1.034m/s
Explanation:
We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,
![m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s](https://tex.z-dn.net/?f=m_1%20%3D%2065000kg%5C%5Cv_1%20%3D%200.8m%2Fs%5C%5Cm_2%20%3D%2092000kg%5C%5Cv_2%20%3D%201.2m%2Fs)
<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,
![V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}](https://tex.z-dn.net/?f=V_%7Bcm%7D%20%3D%20%5Cfrac%7Bm_1v_1%2Bm_2v_2%7D%7Bm_1%2Bm_2%7D)
Substituting,
![V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}](https://tex.z-dn.net/?f=V_%7Bcm%7D%20%3D%20%5Cfrac%7B%2865000%2A0.8%29%2B%2892000%2A1.2%29%7D%7B92000%2B65000%7D)
![V_{cm} = 1.034m/s](https://tex.z-dn.net/?f=V_%7Bcm%7D%20%3D%201.034m%2Fs)
Part B)
For the Part B we need to apply conserving momentum equation, this formula is given by,
![m_1v_1+m_2v_2 = (m_1+m_2)v_f](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%20%3D%20%28m_1%2Bm_2%29v_f)
Where here
is the velocity after the collision.
![v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Cfrac%7Bm_1v_1%2Bm_2v_2%7D%7Bm_1%2Bm_2%7D)
![v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}](https://tex.z-dn.net/?f=v_f%20%3D%20%5Cfrac%7B%2865000%2A0.8%29%2B%2892000%2A1.2%29%7D%7B92000%2B65000%7D)
![v_f = 1.034m/s](https://tex.z-dn.net/?f=v_f%20%3D%201.034m%2Fs)
When an object absorbs an amount of energy equal to Q, its temperature raises by
![\Delta T](https://tex.z-dn.net/?f=%5CDelta%20T)
following the formula
![Q=m C_s \Delta T](https://tex.z-dn.net/?f=Q%3Dm%20C_s%20%5CDelta%20T)
where m is the mass of the object and
![C_s](https://tex.z-dn.net/?f=C_s%20)
is the specific heat capacity of the material.
In our problem, we have
![Q=2.44 \cdot 10^3 J](https://tex.z-dn.net/?f=Q%3D2.44%20%5Ccdot%2010%5E3%20J)
,
![m=235.0 g](https://tex.z-dn.net/?f=m%3D235.0%20g)
and
![\Delta T=35 K](https://tex.z-dn.net/?f=%5CDelta%20T%3D35%20K)
, so we can re-arrange the formula and substitute the numbers to find the specific heat capacity of the metal:
The answer is c to take a pictures with the camera in the metal hall
Answer: 0.817A
Explanation:
Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It
It means a current of one A flow in the opposite direction.
This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,
In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.
According to this, the current in the given problem is
[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s
= 0.817 A