Answer:
α=9.42 rad/s T= 0.014 rad/s
Explanation:
Low speed= Wi=3 rps = 3*2*π rad/s= 6π rad/sec
Medium speed= Wf= 9 rps= 9*2*π rad/sec= 18 rad/sec
time= t= 4 sec
Angular acceleration=α=
⇒α=(18 π - 6 π)/4
⇒α=9.42 rad/s
b).
Rotational inertia= I = 1.5*10^-3 kg-m^2
Net torque= T = I*α = (1.5*10^-3) * (9.42) rad/s^2
⇒T= 0.014 rad/s^2
Answer:
Explanation:
Resonant frequency is 240
4π² x 240² = 1 / LC
230400π² = 1 / LC
Let the required frequency = n
inductive reactance = 2 πn L
capacitative reactance = 1 / 2 π n C
inductive reactance / capacitative reactance
= 4π² x n ² x LC = 5.68
4π² x n ² = 1 / LC x 5.68
= 230400π² x 5.68
4n ²= 230400 x 5.68
n ²= 57600 x 5.68
n ² = 327168
n = 572 approx
Answer:
Explanation:
Considering Work done by friction is asked
Given
mass of block 
force 
inclination 
block is displaced by 
coefficient of kinetic friction 
Friction force 
Normal reaction 
Friction Force 
Work done by friction force
the same with that of products
Explanation:
In a chemical reaction, the total charge of the reactants must be the same with that of products.
Charges must be conserved or balanced in chemical reactions.
- In both acidic and basic/neutral medium electrons are used to balance the charge.
- The appropriate number of electrons is added to the side with a larger charge.
- One electron is used to balance each positive charge.
- This ensures that the sum of charges on both sides the same.
Learn more:
Balanced equation brainly.com/question/5297242
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Answer:
b) True. the force of air drag on him is equal to his weight.
Explanation:
Let us propose the solution of the problem in order to analyze the given statements.
The problem must be solved with Newton's second law.
When he jumps off the plane
fr - w = ma
Where the friction force has some form of type.
fr = G v + H v²
Let's replace
(G v + H v²) - mg = m dv / dt
We can see that the friction force increases as the speed increases
At the equilibrium point
fr - w = 0
fr = mg
(G v + H v2) = mg
For low speeds the quadratic depended is not important, so we can reduce the equation to
G v = mg
v = mg / G
This is the terminal speed.
Now let's analyze the claims
a) False is g between the friction force constant
b) True.
c) False. It is equal to the weight
d) False. In the terminal speed the acceleration is zero
e) False. The friction force is equal to the weight
