Answer:
-414.96 N
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration


The force the ground exerts on the parachutist is -414.96 N
If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase
Answer:
266 g or 0.266 kg
Explanation:
The formula for specific heat capacity is given as,
Q = cm(t₂-t₁) ..................... Equation 1
Where Q = Heat Energy, c = specific heat capacity of Aluminum, m = mass of the aluminum fins, t₁ = initial temperature, t₂ = final temperature.
make m the subject of the equation,
m = Q/c(t₂-t₁)................... Equation 2
Given : Q = 2571 J, c = 0.897 J/g.°C, t₁ = 15.73 °C, t₂ = 26.50 °C.
Substitute into equation 2
m = 2571/[0.897×(26.5-15.73)]
m = 2571/9.661
m = 266 g or 0.266 kg
Hence the mass of the Aluminum fins = 266 g or 0.266 kg
Answer:
work done on desk = m g h = 105 * 9.81 * 2.46 = 2534 Joules
Explanation:
so work in = 2534 / 0.875 = 2896 Joules
that is your 648 times distance your hand moved holding the rope
2896 = 648 * x
4.47 meters
I think maybe a typo, 648 N is too big, maybe 64.8 ? Any block and tackle system does better than that.
Answer:
Keq = 2k₃
Explanation:
We can solve this exercise using Newton's second one
F = m a
Where F is the eleatic force of the spring F = - k x
Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement
F + F = m a
k₃ x + k₃ x = m a
a = 2k₃ x / m
To find the effective force constant, suppose we change this spring to what creates the cuddly displacement
Keq = 2k₃
Answer:
ε₂ =2.63 V
Explanation:
given,
M = 0.0034 H
I (t) = I₀ sin (ωt)
I (t) = 5.4 sin (143 t)


magnitude of the induced emf in the second coil
ε₂ =
ε₂ =
for maximum emf
cos (143 t) = 1
ε₂ =
ε₂ =2.63 V