C. inertia. the man is sent flying off the bus because of his weight and the sudden stop of the bus. this effect is called inertia. an example of gravity would be throwing an apple up and having it come to the ground. an example of weight would be putting a man and an elephant on a scale and having the elephant come down while the man goes up.
Answer:
A) 35 ft
B) 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Explanation:
A) Total distance covered by the dog = 20 + 15
= 35 ft
B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;
total displacement of the dog = 20 - 15
= 5 ft
C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick
But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.
Thus,
Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick
Answer:
27.22 m/s
Explanation:
Let the speed of clay before impact is u.
the speed of clay and target is v after impact.
use conservation of momentum
momentum before impact momentum after impact
mass of clay x u = (mass of clay + mass of target) x v
100 x u = (100 + 500) x v
u = 6 v .....(1)
distance, s = 2.1 m
μ = 0.5
final velocity is zero. use third equation of motion
v'² = v² + 2as
0 = v² - 2 x μ x g x s
v² = 2 x 0.5 x 9.8 x 2.1 = 20.58
v = 4.54 m/s
so by equation (1)
u = 6 x 4.54 = 27.22 m/s
thus, the speed of clay before impact is 27.22 m/s.
a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J
b.) K 2 +GmM( r 11− r 21)=6.9×10 7 J
Applying Law of Energy conservation :
K 1+U 1
=K 2+U 2
⇒K 1− r 1GmM
=K 2− r 2 GmM
where M=5.0×10 23kg,r1
=> R=3.0×10 6m and m=10kg
(a) If K 1
=5.0×10 7J and r 2
=4.0×10 6 m, then the above equation leads to
K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J
(b) In this case, we require K 2
=0 and r2
=8.0×10 6m, and solve for K 1:K 1
=K 2 +GmM (r 11− r 21)=6.9×10 7 J
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