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Answer:
(a) Acceleration of the bag will be a=16.214m/sec^2
(B) Weight of the bag will be 137.2 N
Explanation:
We have given mass of the bag m = 14 kg
Force with which bag is lifted = 227 N
(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration
So 
(b) Acceleration due to gravity 
We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity
So weight 
So weight of the bag will be 137.2 N
By compressing the spring a distance <em>x</em> (in m), you are storing 1/2 <em>k</em> <em>x</em> ² (in J) of potential energy, which is converted completely into kinetic energy 1/2 <em>m v</em> ², where
• <em>k</em> = 40 N/m = spring constant
• <em>m</em> = 10 kg = mass of the ball
• <em>v</em> = 2 m/s = ball's speed (at the moment the spring returns to its equilibrium point)
So we have
1/2 <em>k</em> <em>x</em> ² = 1/2 <em>m</em> <em>v</em> ²
<em>x</em> = √(<em>m</em>/<em>k</em> <em>v</em> ²) = √((10 kg) / (40 N/m) (2 m/s)²) = 1 m
The question is incomplete, the complete question is;
A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?
Answer:
See explanation
Explanation:
From Newton's law of cooling;
θ1 - θ2/t = K(θ1 + θ2/2 - θo]
Where;
θ1 and θ2 are initial and final temperatures
θo is the temperature of the surroundings
K is the constant
t is the time taken
Hence;
100 - 60/5 = K(100 + 60/2 - θo)
100 - 40/10 = K(100 + 40/2 - θo)
8= (80 - θo)K -----(1)
6= (70 - θo)K -----(2)
Diving (1) by (2)
8/6 = (80 - θo)/(70 - θo)
8(70 - θo) = 6(80 - θo)
560 - 8θo = 480 - θo
560 - 480 = -θo + 8θo
80 = 7θo
θo = 11.4°
Again from Newton's law of cooling;
θ = θo + Ce^-kt
Where;
t= 0, θ = 60° and θo = 11.4°
60 = 11.4 + C e^-K(0)
60 - 11.4 = C
C= 48.6°
To obtain K
40 = 11.4 + 48.6e^-10k
40 -11.4 = 48.6e^-10k
28.6/48.6 = e^-10k
0.5585 = e^-10k
-10k = ln0.5585
k= ln0.5585/-10
K= 0.0583
Hence, the temperature in 15 minutes;
θ= 11.4 + 48.6e^(-0.0583 × 15)
θ= 31.7°
