Asked by Fate
A 15kg wedding cake is allowed to slide freely down a smooth 30degree inclined. Find the resultant force down the inclined and the acceleration of the object.
Answers
Answered by
Henry
Fc=mg = 15kg * 9.8N/kg = 147N @ 30Deg = weight of cake.
1. Fp = 147sin30 = 73.5N = Force parallel to the plane downward.
2. F = ma.
a = F/m = 73.5 / 15 = 4.9m/s^2
1. Fp = 147sin30 = 73.5N = Force parallel to the plane downward.
2. F = ma.
a = F/m = 73.5 / 15 = 4.9m/s^2
Answered by
Theresia
Pls help me with this question
Answered by
Faile
Fr=Fg sin 30 degree
=masin 30
=15kg×10m/s^2×Sin 30
=150 Sin 30
=75N
Acceleration
F= ma
a= F/m
=75N /15kg
= 5 m/s ^2
=masin 30
=15kg×10m/s^2×Sin 30
=150 Sin 30
=75N
Acceleration
F= ma
a= F/m
=75N /15kg
= 5 m/s ^2
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