Answer:
78.4 KN/m
Explanation:
Given
mass of person 'm' =80 kg
car dips about i.e spring stretched 'x'= 1 cm => 0.01m
acceleration due to gravity 'g'= 9.8 m/s^2
as we know that,in order to find approximate spring constant we use Hooke's Law i.e F=kx
where,
F = the force needed
x= distance the spring is stretched or compressed beyond its natural length
k= constant of proportionality called the spring constant.
F=kx ---> (since f=mg)
mg=kx
k=(mg)/x
k=(80 x 9.8)/ 0.01
k=78.4x
k=78.4 KN/m
Explanation:
Given that,
Initial speed of the bag, u = 7.3 m/s
Height above ground, s = 24 m
We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :
v = 22.88 m/s
So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.
Answer:
205 V
V = 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is
w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V = IR
= (0.044 A) (93 Ω)
V = 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V = V
cos (wt)
Putting V = 4.092 V and w = 500 rad/s
V = V
cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V = 2.05 V