Here is what we know:
a = ?, s = ?, u = 18.1m/s, v = zero/rest, t = 7.0s, m = 2110kg
(a = acceleration, s= displacement, u = initial velocity, v = final velocity, t = time and m is mass)
Now we choose a kinematic formula. Since we know v, u and t, we will use the formula: v = u+at and rearrange it so that we can find a.
a = v-u/t
a = 0-18.1/7.0 = -2.5857...
therefore, a = -2.6m/s
We have our acceleration, now let’s find the net force. To find the force we use one of Newtons laws of motion.
We will use Newtons second law since it describes what happens when one or more forces act upon an object.
F = ma
F = (2110kg)(-2.6m/s)
F = -5486 kg
Therefore the net force F = -5486 N
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Answer:
It's not what happens to you. It's A.) how you face it
Explanation:
Answer:
V = 48.49m/s
Explanation:
Given the following information:
Combined mass = 92kg
Hill's height = 120m
Course of ride = 384m
Frictional force = 261N
Initial speed (u) = 9m/s
Final speed (v) = ?
Since we are looking for her speed at the bottom,
Time = distance/speed = 384m/9m.s
Time = 42.67s
we use the equation
H = V²/2g ( equation for maximum heigh of trajectory)
Therefore, plugging the values we have
120 = V²/2×9.8
V² = 9.8×120×2
V = √2352
V = 48.49m/s
