Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km
3.0 A i got it off Quizlet and there usually always right lol can't submit tho my answers to short.... Dot dot dot
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
The light will bend when in
Answer:
2.75 m/s^2
Explanation:
The airplane's acceleration on the runway was 2.75 m/s^2
We can find the acceleration by using the equation: a = (v-u)/t
where a is acceleration, v is final velocity, u is initial velocity, and t is time.
In this case, v is 71 m/s, u is 0 m/s, and t is 26.1 s Therefore: a = (71-0)/26.1
a = 2.75 m/s^2
