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steposvetlana [31]
2 years ago
6

Which rain forest is located mostly on islands

Physics
2 answers:
amm18122 years ago
7 0
Tropical Rain forests are mostly located around the equator found from South America to Africa.
natita [175]2 years ago
6 0
The tropical rain forest is located mostly on Islands. Mainly found in Africa.
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Water flows at 10 m/s through a pipe with radius 0.025 m. The pipe goes up to the second floor of the building, 2.5 m higher, an
Lera25 [3.4K]

Answer: from the information given, the velocity of the water will decrease but the pipe size will remain the same.

This can be proved with bernoulli's equation.

Explanation: careful analysis of the system using bernoulli's equation of flow is shown in the image attached

5 0
3 years ago
The relationship between sharks and remoras is characterized<br><br>TRUE <br>   OR<br>FALSE
Damm [24]
False, sharks and remoras have a symbiotic relationship. The remora removes parasites from the sharks scales, and the shark provides protection for the remora
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3 years ago
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
charle [14.2K]

Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

Height = 0.15 m

Radius of circular arc = 0.27 m

Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

Calculate the total energy of the composite system at any time after the collision.

We need to calculate the total energy of the composite system

Total energy of the system at any time = Potential energy of the system at the stopping point

E=mgh+Mgh

E=(m+M)gh

Put the value in to the formula

E=(30\times10^{-3}+5.29)\times 9.8\times0.15

E=7.8\ J

Hence, The total energy of the composite system is 7.8 J.

8 0
3 years ago
A solid circular disk has a mass of 1.2 kg and a radius of 0.16m. Each of three identical thin rods has a mass of 0.16kg. The ro
Juli2301 [7.4K]

Answer:

0.027648 kgm²

Explanation:

M = Mass of disc = 1.2 kg

r = Radius of disc = 0.16 m

m = Mass of rod = 0.16 kg

R = Rod distance = 0.16 m

Moment of inertia of disk is given by

I_1=\dfrac{1}{2}Mr^2\\\Rightarrow I_1=\dfrac{1}{2}1.2\times 0.16^2\\\Rightarrow I_1=0.01536\ kgm^2

Moment of inertia of the three rods

I_2=3mr^2\\\Rightarrow I_2=3\times 0.16\times 0.16^2\\\Rightarrow I_2=0.012288\ kgm^2

The total moment of inertia is given by

I=I_1+I_2=0.01536+0.012288\\\Rightarrow I=0.027648\ kgm^2

The moment of inertia of the stool with respect to an axis that is perpendicular to the plane of the disk at its center is 0.027648 kgm²

8 0
3 years ago
A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction
nikitadnepr [17]

Answer:

71.76 m

Explanation:

We will solve this question using the work energy theorem.

The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.

ΔK.E = W

In the attached free body diagram for the question, the forces acting on the puck are given.

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J (since the puck comes to a stop)

Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J

ΔK.E = 0 - 67.6 = - 67.6 J

W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)

Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h

Workdone by the frictional force = F × d

F = μ N

μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)

N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N

F = μ N = 0.3 × 1.697 = 0.509 N

where d = distance along the incline that the puck travels.

d = h/sin 30° = 2h (from trigonometric relations)

Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h

ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)

- 67.6 = - 1.96h + 1.02h

-0.942h = - 67.6

h = 71.76 m

6 0
2 years ago
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