Expand each vector into their component forms:
![\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20A%3D%284.5%5C%2C%5Cmathrm%20N%29%28%5Ccos%5Ctheta_A%5C%2C%5Cvec%5Cimath%2B%5Csin%5Ctheta_A%5C%2C%5Cvec%5Cjmath%29%3D%282.58%5C%2C%5Cvec%5Cimath%2B3.69%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
Similarly,
![\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20B%3D%28-1.23%5C%2C%5Cvec%5Cimath%2B0.860%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
![\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20C%3D%28-3.44%5C%2C%5Cvec%5Cimath-4.91%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
Then assuming the resultant vector
is the sum of these three vectors, we have
![\vec R=\vec A+\vec B+\vec C](https://tex.z-dn.net/?f=%5Cvec%20R%3D%5Cvec%20A%2B%5Cvec%20B%2B%5Cvec%20C)
![\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N](https://tex.z-dn.net/?f=%5Cvec%20R%3D%28-2.09%5C%2C%5Cvec%5Cimath-0.368%5C%2C%5Cvec%5Cjmath%29%5C%2C%5Cmathrm%20N)
and so
has magnitude
![\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20R%5C%7C%3D%5Csqrt%7B%28-2.09%29%5E2%2B%28-0.368%29%5E2%7D%5C%2C%5Cmathrm%20N%5Capprox2.12%5C%2C%5Cmathrm%20N)
and direction
such that
![\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ](https://tex.z-dn.net/?f=%5Ctan%5Ctheta_R%3D%5Cdfrac%7B-0.368%7D%7B-2.09%7D%5Cimplies%5Ctheta_R%3D-170%5E%5Ccirc%3D190%5E%5Ccirc)
With constant angular acceleration
, the disk achieves an angular velocity
at time
according to
![\omega=\alpha t](https://tex.z-dn.net/?f=%5Comega%3D%5Calpha%20t)
and angular displacement
according to
![\theta=\dfrac12\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac12%5Calpha%20t%5E2)
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
![21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}](https://tex.z-dn.net/?f=21.0%5C%2C%5Cmathrm%7Brad%7D%3D%5Cdfrac12%5Calpha%281.00%5C%2C%5Cmathrm%20s%29%5E2%5Cimplies%5Calpha%3D42.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D)
b. Under constant acceleration, the average angular velocity is equivalent to
![\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5Comega_f%2B%5Comega_i%7D2)
where
and
are the final and initial angular velocities, respectively. Then
![\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega_%7B%5Crm%20avg%7D%3D%5Cdfrac%7B%5Cleft%2842.0%5Cfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%7D2%3D42.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
c. After 1.00 s, the disk has instantaneous angular velocity
![\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}](https://tex.z-dn.net/?f=%5Comega%3D%5Cleft%2842.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%3D42.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D)
d. During the next 1.00 s, the disk will start moving with the angular velocity
equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle
according to
![\theta=\omega_0t+\dfrac12\alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%3D%5Comega_0t%2B%5Cdfrac12%5Calpha%20t%5E2)
which would be equal to
![\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cleft%2842.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Crm%20s%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%2B%5Cdfrac12%5Cleft%2842.0%5Cdfrac%7B%5Crm%20rad%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29%281.00%5C%2C%5Cmathrm%20s%29%5E2%3D63.0%5C%2C%5Cmathrm%7Brad%7D)
Answer: 2. Solution A attains a higher temperature.
Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.
In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.
Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.
<em>We have a formula for such condition,</em>
.....................................(1)
where:
= temperature difference
- c= specific heat of the body
<u>Proving mathematically:</u>
<em>According to the given conditions</em>
- we have equal masses of two solutions A & B, i.e.
![m_A=m_B](https://tex.z-dn.net/?f=m_A%3Dm_B)
- equal heat is supplied to both the solutions, i.e.
![Q_A=Q_B](https://tex.z-dn.net/?f=Q_A%3DQ_B)
- specific heat of solution A,
![c_{A}=2.0 J.g^{-1} .\degree C^{-1}](https://tex.z-dn.net/?f=c_%7BA%7D%3D2.0%20J.g%5E%7B-1%7D%20.%5Cdegree%20C%5E%7B-1%7D)
- specific heat of solution B,
![c_{B}=3.8 J.g^{-1} .\degree C^{-1}](https://tex.z-dn.net/?f=c_%7BB%7D%3D3.8%20J.g%5E%7B-1%7D%20.%5Cdegree%20C%5E%7B-1%7D)
&
are the change in temperatures of the respective solutions.
Now, putting the above values
![Q_A=Q_B](https://tex.z-dn.net/?f=Q_A%3DQ_B)
![m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1](https://tex.z-dn.net/?f=m_A.c_A.%20%5CDelta%20T_A%3Dm_B.c_B%20.%20%5CDelta%20T_B%5C%5C%5C%5C2.0%5Ctimes%20%5CDelta%20T_A%3D3.8%20%5Ctimes%20%5CDelta%20T_B%5C%5C%5C%5C%20%5CDelta%20T_A%3D%5Cfrac%7B3.8%7D%7B2.0%7D%5Ctimes%20%5CDelta%20T_B%5C%5C%5C%5C%5C%5C%5Cfrac%7B%5CDelta%20T_%7BA%7D%7D%7B%5CDelta%20T_%7BB%7D%7D%20%3D%20%5Cfrac%7B3.8%7D%7B2.0%7D%3E1)
Which proves that solution A attains a higher temperature than solution B.
Answer:
300 cos 30 = 40 a + 40 * .2 * 10
Total force = mass * acceleration + frictional force
260 = 40 a + 80
a = 180 / 40 = 4.5 m/s^2
Check:
15 a + 15 * 10 * .2 = T acceleration of 15 kg block (assuming a = 4.5)
T = 15 (4.5) + 30 = 97.5 force required to accelerate 15 kg block
260 - 97.5 = 162.5 net force on 25 kg block
162.5 = 4.5 (25) + 25 * 10 * .2
162.5 = 112.5 + 50 = 162.5
4.5 m/s^2 checks out as correct