Answer:
distance when the weight is 8 kg is 26.66 cm
Explanation:
given data
distance d2 = 10 cm
weight w2 = 3 kg
weight w1 = 8 kg
to find out
distance when the weight is 8 kg
solution
we consider here distance d1 when weight is 8 kg
so equation will be
d1/d2 = w1/w2
d/10 = 8/ 3
so d = 8/3 × 10
so d = 26.66
distance when the weight is 8 kg is 26.66 cm
Answer:
According to the parallelogram law of vector addition if two vectors act along two adjacent sides of a parallelogram(having magnitude equal to the length of the sides) both pointing away from the common vertex, then the resultant is represented by the diagonal of the parallelogram passing through the same common vertex
Explanation:
Answer:
1.8 m/s²
36 N
34.8 N
Explanation:
For the monkey :
m₁ = mass of monkey = 4.50 kg
T₁ = Tension force in the rope on monkey's side
a = acceleration
From the force diagram, force equation for the motion of monkey is given as
m₁ g - T₁ = m₁ a
(4.50 x 9.8) - T₁ = 4.5 a
T₁ = 44.1 - 4.5 a eq-1
For the bunch of bananas :
m₂ = mass of bunch of bananas = 3 kg
T₂ = tension force in the rope on the side of banana
From the force diagram, force equation for the motion of bananas is given as
T₂ - m₂ g = m₂ a
T₂ - (3 x 9.8) = 3 a
T₂ = 29.4 + 3 a eq-2
m = mass of the pulley = 1.50 kg
r = radius of the pulley = 0.090 m
α = angular acceleration of pulley = a/r
Torque equation for the pulley is given as
(T₁ - T₂ )r = I α
(T₁ - T₂ )r = I (a/r)
T₁ - T₂ = (0.5 m r²) (a/r²)
T₁ - T₂ = (0.5) ma
using eq-1 and eq-2
44.1 - 4.5 a - (29.4 + 3 a) = (0.5) ma
44.1 - 4.5 a - (29.4 + 3 a) = (0.5) (1.50) a
a = 1.8 m/s²
Using eq-1
T₁ = 44.1 - 4.5 a
T₁ = 44.1 - 4.5 (1.8)
T₁ = 36 N
using eq-2
T₂ = 29.4 + 3 a
T₂ = 29.4 + 3 (1.8)
T₂ = 34.8 N
Answer:
the answer is 0 .75m
Explanation:
where are finding the distance the speed is given as well as the time..so you multiple the speed with the time thus you get the distance..i hope that's the answer..
Answer:
35.5 g of chlorine would be in the compound
Explanation:
using the mole concept
first write down your balanced chemical equation
CaCl2_____Ca +Cl2
then if 40g of Ca____ 71 g of cl
so 20g of Ca_____Xg of cl
cross multiplying we have
X= (20 *71)/40 = 35.5g