Answer:
The dog catches up with the man 6.1714m later.
Explanation:
The first thing to take into account is the speed formula. It is 
, where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is 
Now, the distance equation for the man would be:
The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.
For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:
Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.
The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.
That value is used in the man's distance equation.
Finally, the dog catches up with the man 6.1714m later.
Answer:
9.47 rad/s^2
Explanation:
Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,
s = 35.4 m
let a be the linear acceleration.
Use III equation of motion.
v^2 = u^2 + 2 a s
7.1 x 7.1 = 0 + 2 x a x 35.4
a = 0.71 m/s^2
Now the relation between linear acceleration and angular acceleration is
a = r x α
where, α is angular acceleration
α = 0.71 / 0.075 = 9.47 rad/s^2
Answer:
Explanation:
An inelastic collision is one where 2 masses collide and stick together, moving as a single mass after the collision occurs. When we talk about this type of momentum conservation, the momentum is conserved always, but the kinetic momentum is not (the velocity changes when they collide). Because there is direction involved here, we use vector addition. The picture before the collision has the truck at a mass of 3520 kg moving north at a velocity of 18.5. The truck's momentum, then, is 3520(18.5) = 65100 kgm/s; coming at this truck is a car of mass 1480 kg traveling east at an unknown velocity. The car's momentum, then, is 1480v. The resulting vector (found when you pick up the car vector and stick the initial end of it to the terminal end of the truck's momentum vector) forms the hypotenuse of a right triangle where one leg is 65100 kgm/s, and the other leg is 1480v. Since we already know the final velocity of the 2 masses after the collision, we can use that to find the final momentum, which will serve as the resultant momentum vector in our equation (we'll get there in a sec). The final momentum of this collision is
p = mv and
p = (3520 + 1480)(13.6) so
p = 68000. Final momentum. The equation for this is a take-off of Pythagorean's Theorem and the one used to find the final magnitude of a resultant vector when you first began your vector math in physics. The equation is
which, in words, is
the final momentum after the collision is equal to the square root of the truck's momentum squared plus the car's momentum squared. Filling in:
and
and
and
and
so
v = 13.3 m/s at 72.6°
Winds that blow from the north and south poles would be called k<span>atabatic winds. I'm not sure if I spelled that right, but that's the answer I hope.</span>
