Answer:
the jumper's speed just as he leaves the ground = 4.2 m/s
the force he must exert on the ground to perform the 0.90 m jump = 4.99 × 10³ N
Explanation:
Given that :
height h = 0.90 m
mass = 51 kg
distance s = 0.10 m
What is the jumper's speed just as he leaves the ground?
The velocity P.E = K.E
where;
g = 9.8
v = 4.2 m/s
the jumper's speed just as he leaves the ground = 4.2 m/s
What force must he exert on the ground to perform the 0.90 m jump?
Let's first determine the acceleration by using the equation of motion;
v² = u² + 2as
4.2² = 0 + 2(a)(0.10)
17.64 = 0.2 a
a = 17.64/0.2
a = 88.2 m/s²
The force F is now calculated by the relation:
F - mg = ma
F = mg+ ma
F = m(g+a)
F = 51(9.8 + 88.2)
F = 4988 N
F = 4.99 × 10³ N
In gases, temperature relates to its kinetic energy — the hotter the gas, the faster the particles will move (for example, if we take that the formula of kinetic energy is Ek = 3/2 * k * T where k iz Boltzmanns constant and T is temperature, we see that kinetic energy — which is the energy an object has when its moving — only really depends on the temperature.)
Answer:
Part a)
Moment of inertia of the cylinder is given as
Part B)
Height of the cylinder is of no use here to calculate the inertia
Part C)
Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as
Explanation:
As we know that the inclined plane is of length L = 3 m
and its inclination is given as 25 degree
so we know that acceleration of center of mass of the cylinder is constant so we will have
so we have
now we know that
Now we have know that final speed of the cylinder due to pure rolling is given as
Part B)
Height of the cylinder is of no use here to calculate the inertia
Part C)
Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as
45-4753that is your answer
Image<span> formed by a </span>plane mirror is<span> always </span>virtual<span> which means that the light rays </span>do<span> not actually come from the </span>image but<span> upright and these of the same shape and size are the object it </span>is<span> reflecting.</span>