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Simora [160]
3 years ago
5

A 98.1 kg horizontal circular platform rotates freely with no friction about its center at an initial angular velocity of 1.69 r

ad/s . A monkey drops a 9.29 kg bunch of bananas vertically onto the platform. They hit the platform at 45 of its radius from the center, adhere to it there, and continue to rotate with it. Then the monkey, with a mass of 20.3 kg , drops vertically to the edge of the platform, grasps it, and continues to rotate with the platform. Find the angular velocity of the platform with its load. Model the platform as a disk of radius 1.51 m .
Physics
1 answer:
sertanlavr [38]3 years ago
6 0

Answer:

Explanation:

The problem is based on conservation of angular momentum.

Moment of inertia of the disc = 1/2 m R² , m is mass of the disc and R is its radius.

= 1/2 x 98.1 x 1.51²

= 111.84 kg m²

Moment of inertia of disc + moment of inertia of bananas + monkey

= 1/2 x 98.1 x 1.51² + 9.29 x .45 x 1.51 + 20.3 x 1.51² ( moment of inertia of banana and monkey will be equal to mass x radial distance from axis² )

= 111.84 + 6.31 +46.28

= 164.43 kg m²

Now applying law of conservation of angular momentum

= I₁ ω₁ = I₂ω₂

111.84 x 1.69 = 164.43 x ω₂

ω₂ = 1.15 rad / s

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tatyana61 [14]

Answer:

324 people

Explanation:

radius r = diameter / 2 = 45cm /2 = 22.5 cm or 0.225 m

We can calculate the total volume of the log by computing the volume of each cylindrical log:

V = \pi r^2 L = \pi 0.225^2 * 6.5 = 4.6 m^3

So the total volume of all 12 logs is

12 * 4.6 = 55.135 m^3

The weight that the raft can support, is the buoyancy force subtracted by its own weight.

The buoyancy force is basically the weight of the water is displaced, which is water density times volume. The log weight is also log density times its volume

F = F_b - F_w = g\rho_wV - g\rho_rV = gV(\rho_w - \rho_r)

where \rho_w = 1000kg/m^3,\rho_r are the density of water and raft, respectively. But since we have the specific gravity of wood is 0.6. That measn

\rho_r / \rho_w = 0.6

\rho_r = 0.6\rho_w

Therefore \rho_w - \rho_r = \rho_w - 0.6\rho_w = 0.4\rho_w = 0.4*1000 = 400 kg/m^3

Let g = 9.8m/s2. We can now calculate the force that the raft can support

F = gV(\rho_w - \rho_r) = 9.8*55.135*400=216129.2N

Each person has a mass of 68kg, their weight would be

W = 68*g = 68*9.8 = 666.4N

So the maximum number of people that the raft can hold is

F / W = 216129.2 / 666.4 = 324 people

3 0
4 years ago
A 39.7 n object is in free fall. what is the magnitude of the net force which acts on the object? answer in units of n.
Vika [28.1K]
If there is no air resistance, the value would still be the same. So therefore, the answer is 39.7 N.
Supposing that there is an air resistance of 19.8 N.With a 19.8 N air resistance: ∑F = 39.7 N - 19.8 N = 19.9 N would be the net force. The important point here is that the force of gravity equivalent the weight of the object.
6 0
4 years ago
The figure shows a velocity-versus-time graph for a particle moving along the x-axis. At t=0s, assume that x=0m. What is the par
swat32

Using the velocity-time graph, the displacement can be calculated by the area under the velocity-time graph. At 3 seconds the total displacement is then equal to (4)(2) + (4 + 2)*1/2 = 11 m. Assuming that the starting point is at x = 0, then the particle at t=3s is at x=11 m.


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
7 0
3 years ago
What is the kinetic energy of a 1700 kg car traveling at a speed of 30 m/s (â65 mph)? does your answer to part b depend on the c
MArishka [77]
The kinetic energy of an object is given by
K= \frac{1}{2} mv^2
where m is the mass of the object and v its velocity.

For the car in the problem, m=1700 kg and v=30 m/s, so the kinetic energy of the car is
K= \frac{1}{2}mv^2= \frac{1}{2}(1700 kg)(30 m/s)^2=7.65\cdot 10^5 J

and as we can see, yes, the answer depends on the car's mass.
6 0
3 years ago
The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl
Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, m_1=5\ kg

Initial speed of block A, u_1=5\ m/s

Mass of block B, m_2=8\ kg

Initial speed of block B, u_2=0

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0
3 years ago
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