Answer:
Vx = 35 x cos(13deg)
Vy = 35 x sin(13deg) - gt
(g is acceleration due to gravity =~9.8 meter/second^2, t is time in second)
Explanation:
The tiger leaps up, then x and y component of its velocity are:
Vx = Vo x cos(alpha)
Vy = Vo x sin(alpha) - gt
(Vo is tiger's initial velocity, alpha is angle between its leaping direction and horizontal plane)
Hope this helps!
Answer:
T= 8.061N*m
Explanation:
The first thing to do is assume that the force is tangential to the square, so the torque is calculated as:
T = Fr
where F is the force, r the radius.
if we need the maximum torque we need the maximum radius, it means tha the radius is going to be the edge of the square.
Then, r is the distance between the edge and the center, so using the pythagorean theorem, r i equal to:
r = 
r = 0.5374m
Finally, replacing the value of r and F, we get that the maximun torque is:
T = 15N(0.5374m)
T= 8.061N*m
Answer: A haploid cell formed in the female uterus
Answer:
a. 32.67 rad/s² b. 29.4 m/s²
Explanation:
a. The initial angular acceleration of the rod
Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.
So, Iα = WL
mL²α/3 = mgL
dividing through by mL, we have
Lα/3 = g
multiplying both sides by 3, we have
Lα = 3g
dividing both sides by L, we have
α = 3g/L
Substituting the values of the variables, we have
α = 3g/L
= 3 × 9.8 m/s²/0.9 m
= 29.4/0.9 rad/s²
= 32.67 rad/s²
b. The initial linear acceleration of the right end of the rod?
The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²
