The distance at which the man slips is 0.3 m
Newton's Second Law, F = ma, is used to calculate the braking distance. By dividing the mass of the car by the gravitational acceleration, one may determine its weight. The weight of the car multiplied by the coefficient of friction equals the brake force.
Given-
mass of man= 70 kg
frictional coefficient μ=0.02
mass of body thrown= m2 = 3kg
let s be the stopping distance
we know that frictional force = F= μN
=μMg= 0.02 x 70 x 10
=14 N
∴acceleration, a= 14/70 = 0.2 m/s²
now on applying conservation of linear momentum
pi=pf pi=0 (initially at rest)
0=m1v1-m2v2 (v1= velocity of man) (v2=velocity of body= 8m/s
v1= m2v2 /m1= 0.3 m/s
we know,
v²- u² = -2as
0- (0.3) ²= -2 x 0.2 x 5
s= 0.09/0.4 ≈ 0.3 m
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Answer:
F = 4212 N
Explanation:
Given that,
Mass of a car, m = 1300 kg
Speed of car on the road is 9 m/s
Radius of curve, r = 25 m
We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :
So, the force has a magnitude of 4212 N
| Impedance | = √ [R² +(ωL)²]
R² = 6800² = 4.624 x 10⁷
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²
= 2.0884 x 10⁻⁴ f²
| Z | = √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ] = 1.6 x 10⁵
(1.6 x 10⁵)² = (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)
(2.56 x 10¹⁰) - (4.624 x 10⁷) = 2.0884 x 10⁻⁴ f²
Frequency² = (2.56 x 10¹⁰ - 4.624 x 10⁷) / 2.0884 x 10⁻⁴
= 2.555 x 10¹⁰ / 2.0884 x 10⁻⁴
= 1.224 x 10¹⁴
= 122,400 GHz <== my calculation
11.1 MHz <== online impedance calculator
Obviously, I must have picked up some rounding errors
in the course of my calculation.
Answer:
Explanation:
The change is as follows
P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process
3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process
In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁
P₁V₁ = n R T₁ , n is no of moles of gas enclosed.
nRT₁ = P₁V₁
Heat added at constant volume = n Cv ( 3T₁ - T₁)
= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)
= 10/3 x nRT₁
= 10/3x P₁V₁
In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁
Heat added at constant pressure in second case
= n Cp ( 15T₁ - 3T₁)
= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)
= 28 x nRT₁
= 28 P₁V₁
