Answer:
Average speed = 46.67 m/s
Explanation:
Given that the time taken in covering first 1000 m = 25 seconds.
The time taken in covering next 2.5 km = 50 seconds.
Total distance covered = 1000 m + 2500 m = 3500 m
Total time taken = 25+50=75 seconds
Average speed = Total distance covered / total time taken
= 3500/75 = 46.67 m/s
Answer:
U = √Rg/sin2θ
Explanation:
Using the formula for "range" in projectile motion to derive the average speed before the ball hits the ground.
Range is the distance covered by the body in the horizontal direction from the point of launch to the point of landing.
According to the range formula,
R = U²sin2θ/g
Cross multiplying we have;
Rg = U²sin2θ
Dividing both sides by sin2θ, we have;
U² = Rg/sin2θ
Taking the square root of both sides we have;
√U² = √Rg/sin2θ
U = √Rg/sin2θ
Therefore, his average speed if he is to meet the ball just before it hits the ground is √Rg/sin2θ
Answer:
t = 3.516 s
Explanation:
The most useful kinematic formula would be the velocity of the motorcylce as a function of time, which is:
Where v_0 is the initial velocity and a is the acceleration. However the problem states that the motorcyle start at rest therefore v_0 = 0
If we want to know the time it takes to achieve that speed, we first need to convert units from km/h to m/s.
This can be done knowing that
1 km = 1000 m
1 h = 3600 s
Therefore
1 km/h = (1000/3600) m/s = 0.2777... m/s
100 km/h = 27.777... m/s
Now we are looking for the time t, for which v(t) = 27.77 m/s. That is:
27.777 m/s = 7.9 m/s^2 t
Solving for t
t = (27.7777 / 7.9) s = 3.516 s
Answer:132.0285
Explanation: Hope this helps!
Answer:
don't know her boi I'm finna glow october
