Answer:
v = -1.8t+36
20 seconds
360 m
40 seconds
36 m/s
The object speed will increase when it is coming down from its highest height.
Explanation:
Differentiating with respect to time we get
a) Velocity of the object after t seconds is v = -1.8t+36
At the highest point v will be 0
b) The object will reach the highest point after 20 seconds
c) Highest point the object will reach is 360 m
d) Time taken to strike the ground would be 20+20 = 40 seconds
Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of
e) The velocity with which the object strikes the ground will be 36 m/s
f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.
Answer:
Fault lines
Explanation:
Earthquakes are most likely to occur near or on fault lines. A great example of this is the ring of fire, a gigantic fault line that gives catastrophic earthquakes.
Answer:
14.36m/s
Explanation:
From the law of conservation of linear momentum
m1u1 + m2u2 = v(m1 + m2)
68×17 + 76×12= v(68+76)
1156+912 = 144v
2068 = 144v
v = 2068/144
=14.36 m/s
Answer:
vf = 3.27[m/s]
Explanation:
In order to solve this problem we must analyze each body individually and find the respective equations. The free body diagram of each body (box and bucket) should be made, in the attached image we can see the free body diagrams and the respective equations.
With the first free body diagram, we determine that the tension T should be equal to the product of the mass of the box by the acceleration of this.
With the second free body diagram we determine another equation that relates the tension to the acceleration of the bucket and the mass of the bucket.
Then we equalize the two stress equations and we can clear the acceleration.
a = 3.58 [m/s^2]
As we know that the bucket descends 1.5 [m], this same distance is traveled by the box, as they are connected by the same rope.
And the speed can be calculated as follows:
Answer:
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