If the two waves combine to produce ANY wave that smaller
than either of the originals, that's destructive interference.
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:
Object 3 has greatest acceleration.
Explanation:
Objects Mass Force
1 10 kg 4 N
2 100 grams 20 N
3 10 grams 4 N
4 1 kg 20 N
Acceleration of object 1,
![a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2](https://tex.z-dn.net/?f=a_1%3D%5Cdfrac%7BF_1%7D%7Bm_1%7D%5C%5C%5C%5Ca_1%3D%5Cdfrac%7B4%7D%7B10%7D%5C%5C%5C%5Ca_1%3D0.4%5C%20m%2Fs%5E2)
Acceleration of object 2,
![a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2](https://tex.z-dn.net/?f=a_2%3D%5Cdfrac%7BF_2%7D%7Bm_2%7D%5C%5C%5C%5Ca_2%3D%5Cdfrac%7B20%7D%7B0.1%7D%5C%5C%5C%5Ca_2%3D200%5C%20m%2Fs%5E2)
Acceleration of object 3,
![a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2](https://tex.z-dn.net/?f=a_3%3D%5Cdfrac%7BF_3%7D%7Bm_3%7D%5C%5C%5C%5Ca_3%3D%5Cdfrac%7B4%7D%7B0.01%7D%5C%5C%5C%5Ca_3%3D400%5C%20m%2Fs%5E2)
Acceleration of object 4,
![a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2](https://tex.z-dn.net/?f=a_4%3D%5Cdfrac%7BF_4%7D%7Bm_4%7D%5C%5C%5C%5Ca_4%3D%5Cdfrac%7B20%7D%7B1%7D%5C%5C%5C%5Ca_3%3D20%5C%20m%2Fs%5E2)
It is clear that the acceleration of object 3 is
and it is greatest of all. So, the correct option is (3).
<span>Answer:
Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by
h = 0.5*g*T^2
However, the speed of the package-rocket system must be sufficient to cross the river in that time
v2 = L/T
Conservation of momentum says that
m1*v1 = (m1 + m2)*v2
where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system.
Expressing v2 in terms of v1
v2 = m1*v1/(m1 + m2)
and then expressing the time in terms of v1
T = (m1 + m2)*L/(m1*v1)
substituting T in the first expression
h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2
solving for v1, the speed before impact is given by
v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1</span>
Answer:
the ball is rolling 7m/s
Explanation:
Formula for kinetic energy: 1/2mv²
K = 1/2mv²
98 = 1/2(4)v²
98 = 2v²
49 = v²
√49 = v
7 = v