Air is being blown into a spherical balloon at the rate of 1.68 in.3/s. Determine the rate at which the radius of the balloon is
1 answer:
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Answer:
B. 2 m/s
B. Acceleration = 4.05 m/s² and Tension = 297.5 N.
Explanation:
A force is applied on a mass m whose acceleration is 4 m/s
Force = mass × acceleration
a = F/m = 4 m/s
4 m/s = F/m
F = 4 m/s (m)
If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:
2F/4 m = F/ 2m
4m/s (m) / 2m = 2 m/s
a = 2 m/s
2.
Given that
mass = 30 kg
mass = 50 kg
= 0.1
From the question; we can arrive at two cases;
That :
----- equation (1)
---- equation (2)
50 a = 50 g - T
30 a = T - 30 g sin 30 - 4 × 30 g cos 30
By summation
80 a =g
80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)
80 a = 324 m/s ²
a = 324/80
a = 4.05 m/s²
From equation , replace a with 4.05
50 × 4.05 = 50 × 10 - T
T = 500 -202.5
T =297.5 N
Answer:
The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)
Explanation:
Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.
Step 2: I must calculate the magnitude of the forces acting on the third charge.
F13: Force exerted by charge 1 on charge 3.
F23: Force exerted by charge 2 on charge 3.
K: Constant of Coulomb's law.
d13: distance from charge 1 to charge 3.
d23: distance from charge 2 to charge 3
Fr: Resulting force.
q1=+2.06 x 10-9 C
q2= -3.27 x 10-9 C
q3= +1.05 x 10-9 C
K=9-10^9 N-m^2/C^2
d13= 0,20 m
d23= 0,10 m
F13= K * (q1 * q3)/(d13)^2
F13=9,7335*10^(-8) N
F23=K * (q2 * q3)/(d23)^2
F23= -3,09 * 10^(-7)
Step 3: We calculate the resultant force on charge 3.
Fr=F13+F23= -2,11665 * 10^(-7)
