A strong gust of wind blows an apple off a tree. The apple has a mass of 0.25 kg. The gust of wind pushes the apple with a force
of 1.05 N to the right. What is the net force on the apple?
5.5 N
- 2.2 N
1.78 N
2.66 N
2 answers:
For this case we must draw the free body diagram of the apple.
It is observed in the attached figure that we have the force of the wind acting to the right, in addition to the force of the weight that goes down.
: It is the force exerted by the weight of the apple (M: It is the mass and g: It is the acceleration of gravity)
It is the force exerted by the wind
Then, the net force will be given by:
We have to:
Substituting we have:
Thus, the net force acting on the apple is 2.66N
Answer:
Option D
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Answer:
Acceleration is 7.990487515m/s²
Initial velocity is 0m.s
Explanation:
s=ut+(1/2)at²
210=0(7.25)+(1/2)a(7.25²)
210=26.28125a
∴a=7.990487515m/s²
'Vi' or 'u' is the inital speed. Since it starts from rest, this equals 0.
Answer:
A.) acceleration= 55.6m/s^2
B.) acceleration of table= 5.0m/s^2
C.) More acceleration
Explanation:
A.) 100N/1.8kg= -55.6
B.) 100N÷20kg= 5
C.) Because since the table would have less mass, it would have had to accelerate more
Answer:
the x and y values
Explanation:
because on the table the x is the input and y is the output
Explanation:
A bearing if an angle is measured clockwise from north direction.
e.g Below the bearing of B from A is 025. (3 figures are always given). the bearing of A from B is 205°.
Answer:
(a): The car's relative position to the base of the cliff is x= 32.52m.
(b): The lenght of the car in the ir is tfall= 1.78 sec.
Explanation:
Vo= 0
V= ?
d= 50m
h= 30m
a= 4 m/s²
t= √(2*d/a)
t= 5 sec
V= a*t
V= 20 m/s
Vx= V * cos(24º)
Vx= 18.27 m/s
Vy= V* sin(24º)
Vy= 8.13 m/s
h= Vy*t + g*t²/2
clearing t:
tfall= 1.78 sec (b)
x= Vx * tfall
x= 32.52 m (a)