Given a = 10 cm/s²
u = 0 cm/s
v = 50 cm/s
we know that
v²=u²+2aS
2500=2×10×S
2500÷20 = S
S= 125 cm
The ramp is 125 cm
Answer:
m = 35.98 Kg ≈ 36 Kg
Explanation:
I₀ = 125 kg·m²
R₁ = 1.50 m
ωi = 0.600 rad/s
R₂ = 0.905 m
ωf = 0.800 rad/s
m = ?
We can apply The law of conservation of angular momentum as follows:
Linitial = Lfinal
⇒ Ii*ωi = If*ωf <em>(I)</em>
where
Ii = I₀ + m*R₁² = 125 + m*(1.50)² = 125 + 2.25*m
If = I₀ + m*R₂² = 125 + m*(0.905)² = 125 + 0.819025*m
Now, we using the equation <em>(I) </em>we have
(125 + 2.25*m)*0.600 = (125 + 0.819025*m)*0.800
⇒ m = 35.98 Kg ≈ 36 Kg
Answer:
568.18 N
Explanation:
From the question,
The formula for gravitational potential is given as
Ep = mgh........................ Equation 1
Where Ep = Gravitational potential, m = mass of the diver,h = Height.
But,
W = mg.................... Equation 2
Where W = weight of the diver.
Substitute equation 2 into equation 1
Ep = Wh
Make W the subject of the equation
W = Ep/h................... Equation 3
Given: Ep = 25000 J, h = 44 m
Substitute into equation 3
W = 25000/44
W = 568.18 N.
Hence the weight of the diver = 568.18 N
Explanation:
It is given that,
Velocity in East, 
Velocity in North, 
(a) The resultant velocity is given by :
(b) The width of the river is, d = 80 m
Let t is the time taken by the boat to travel shore to shore. So,
t = 16 seconds
(c) Let x is the distance covered by the boat to reach the opposite shore. So,
x = 48 meters
Hence, this is the required solution.
