Answer:
6.88 mA
Explanation:
Given:
Resistance, R = 594 Ω
Capacitance = 1.3 μF
emf, V = 6.53 V
Time, t = 1 time constant
Now,
The initial current, I₀ = 
or
I₀ = 
or
I₀ = 0.0109 A
also,
I = ![I_0[1-e^{-\frac{t}{\tau}}]](https://tex.z-dn.net/?f=I_0%5B1-e%5E%7B-%5Cfrac%7Bt%7D%7B%5Ctau%7D%7D%5D)
here,
τ = time constant
e = 2.717
on substituting the respective values, we get
I = ![0.0109[1-e^{-\frac{\tau}{\tau}}]](https://tex.z-dn.net/?f=0.0109%5B1-e%5E%7B-%5Cfrac%7B%5Ctau%7D%7B%5Ctau%7D%7D%5D)
or
I = ![0.0109[1-2.717^{-1}]](https://tex.z-dn.net/?f=0.0109%5B1-2.717%5E%7B-1%7D%5D)
or
I = 0.00688 A
or
I = 6.88 mA
Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
Answer:
Explanation:
a )
Reaction force of the ground
R = mg
= 160 N
Maximum friction force possible
= μ x R
= μ x 160
= .4 x 160
= 64 N .
b )
160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,
Taking moment about top point of ladder
160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3
240 + 444 + 4f = 2700
f = 504 N
c )
Let x be the required distance.
Taking moment about top point of ladder
160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )
240 + 444 x + 1440 = 2700
x = 2.3 m
so man can go upto 2.3 at which maximum friction acts .
The answer is number 2 stomata.
