Answer:
<h3>473.8 m/s; 473.8 m/s</h3>
Explanation:
Given the initial velocity U = 670m/s
Horizontal velocity Ux = Ucos theta
Vertical component of the cannon velocity Uy = Usin theta
Given
U = 670m/s
theta = 45°
horizontal component of the cannonball’s velocity = 670 cos 45
horizontal component of the cannonball’s velocity = 670(0.7071)
horizontal component of the cannonball’s velocity = 473.757m/s
Vertical component of the cannonball’s velocity = 670 sin 45
Vertical component of the cannonball’s velocity = 670 (0.7071)
Vertical component of the cannonball’s velocity = 473.757m/s
Hence pair of answer is 473.8 m/s; 473.8 m/s
Answer:
C = 1.01
Explanation:
Given that,
Mass, m = 75 kg
The terminal velocity of the mass, 
Area of cross section, 
We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,
R = W
or
Where
is the density of air = 1.225 kg/m³
C is drag coefficient
So,
So, the drag coefficient is 1.01.
Answer:
about 229 feet.
Explanation:
According to my research on the information provided by the drivers educational book, It is said that a motor vehicle with good brakes that is going at 50 miles per hour can be stopped within about 229 feet. This is dependent 100% on having good brakes as well as there being normal driving conditions (on pavement with no rain or other weather that may affect driving conditions).
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If I remember correctly, it is the 3rd answer choice.
No, because he was a philosopher
