Answer:
t = 7,8 s
Explanation:
From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .
When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit
Then for the cat arrives at 3,9 m/s nedds
v = vo + a*t vo = 0 then v = a*t
3,9 ( m/s) = 0,5 ( m/s² ) * t
t = 7,8 s
v = 3,9 m/s =
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:
The slope of a velocity graph represents the acceleration of the object. So, the value of the slope at a particular time represents the acceleration of the object at that instant.
Since the angle is West of North, therefore to find for
the westward component (horizontal component) of the vector, we use the sin
function:
sin θ = opposite side / hypotenuse = westward component /
resultant vector
So the westward component (x) is:
x = 85.42 sin 23
<span>x = 33.38 unit</span>
Answer:
The tension in the string is equal to Ct
And the time t0 when the rension in the string is 27N is 3.6s.
Explanation:
An approach to solving this problem jnvolves looking at the whole system as one body by drawing an imaginary box around both bodies and taking summation of the forces. This gives F2 - F1 = Ct. This is only possible assuming the string is massless and does not stretch, that way transmitting the force applied across it undiminished.
So T = Ct
When T = 27N then t = T/C = 27/7.5 = 3.6s
