Answer:
approximately 30 degrees
Explanation:
If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component 
of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):
By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:
which tells us that the closer answer shown is 
Answer:
Temperature of the hot reservoir is 1540K
Explanation:
![E= 1- \frac{T_{c}}[tex]{T_h}=308+{T_c}\\Efficiency of a carnot engine is given by the aboveTc=temperature of the cold reservoirTh= temperature of the hot reservoirK=273+ 35 (convert 35°C to kelvin)K=308k{T_h}={T_c}+308-----------------------(equation 1)20%=1-{T_c}/{T_h}](https://tex.z-dn.net/?f=E%3D%201-%20%5Cfrac%7BT_%7Bc%7D%7D%5Btex%5D%7BT_h%7D%3D308%2B%7BT_c%7D%5C%5C%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3EEfficiency%20of%20a%20carnot%20engine%20is%20given%20by%20the%20above%3C%2Fp%3E%3Cp%3ETc%3Dtemperature%20of%20the%20cold%20reservoir%3C%2Fp%3E%3Cp%3ETh%3D%20temperature%20of%20the%20hot%20reservoir%3C%2Fp%3E%3Cp%3EK%3D273%2B%2035%20%20%28convert%20%2035%C2%B0C%20to%20kelvin%29%3C%2Fp%3E%3Cp%3EK%3D308k%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%7BT_h%7D%3D%7BT_c%7D%2B308-----------------------%28equation%20%201%29%3C%2Fstrong%3E%3C%2Fp%3E%3Cp%3E%3Cstrong%3E20%25%3D1-%7BT_c%7D%2F%7BT_h%7D)
0.2=({T_c}+308-{T_c})/{T_c}+308
.2({T_c}+61.6=308
0.2{T_c}=246.4
{T_c}=1232
recall from equation 1
{T_h}=308+1232
{T_h}=1540K
Oi mate 3 electrons will be in tha 2nd energy level you silly willy.
Answer:
The correct option is;
B. Object X travels at -2 m/s and object Y travels at 4 m/s after the spring is no longer compressed
Explanation:
The given parameters are;
The mass of object Y = M
The mass of object X = 2·M
The initial velocity of object X and object Y = 0 m/s
Let A represent the velocity of object X after the spring is released and B represent the velocity of object Y after the spring is released, therefore, by the principle of the conservation of linear momentum, we have;
(M + 2·M) × 0 = M × B + 2·M × A
∴ (M + 2·M) × 0 = 0 = M × B + 2·M × A
M × B = -2·M × A
∴ B = -2·A
Therefore, the velocity of the object Y = -2 × The velocity of the object X
Whereby the velocity of the object X = -2, The velocity of the object Y = -2 × -2 = 4
Which gives, object X travels at -2 m/s and object Y travels at 4 m/s after the spring is no longer compressed.
Answer:
Correct option a. one state variable T.
Explanation:
In the case of an ideal gas it is shown that internal energy depends exclusively on temperature, since in an ideal gas any interaction between the molecules or atoms that constitute it is neglected, so that internal energy is only kinetic energy, which depends Only of the temperature. This fact is known as Joule's law.
The internal energy variation of an ideal gas (monoatomic or diatomic) between two states A and B is calculated by the expression:
ΔUAB = n × Cv × (TB - TA)
Where n is the number of moles and Cv the molar heat capacity at constant volume. Temperatures must be expressed in Kelvin.
An ideal gas will suffer the same variation in internal energy (ΔUAB) as long as its initial temperature is TA and its final temperature TB, according to Joule's Law, whatever the type of process performed.
