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Savatey [412]
3 years ago
13

which would be better suited for agriculture, the soil of a tropical rain forest or that of a temperate deciduous forest

Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
6 0
I think it may be that of a temperate deciduous forest tho im not sure
thank u for letting me answer and god bless have a good life <3
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reviews the approach taken in problems such as this one. A bird watcher meanders through the woods, walking 0.916 km due east, 0
Verizon [17]

Answer:

Displacement: 2.230 km    Average velocity: 1.274\frac{km}{h}

Explanation:

Let's represent displacement by the letter S and the displacement in direction 49.7° as A. Displaement is a vector, so we need to decompose all the bird's displacement into their X-Y compoments. Let's go one by one:

  • 0.916 km due east is an horizontal direction and cane be seen as  direction towards the negative side of X-axis.
  • 0.928 km due south is a vertical direction and can be seen as a direction towards the negative side of Y-axis.
  • 3.52 km in a direction of 49.7° has components on X and Y  axes. It is necessary to break it down using trigonometry,

First of all. We need to sum all the X components and all the Y componets.

∑Sx = Ax -0.916 ⇒  ∑Sx = [tex]3.52cos(49.7) - 0.916

∑Sx = 1.361 km

∑Sy = Ay - 0.918 ⇒ ∑Sy = 3.52sin(49.7) - 0.918

∑Sy = 1.767

The total displacement is calculated using Pythagoeran therorem:

S_{total} =\sqrt{Sx^{2}+ Sy^{2} } ⇒

S_{total} = 2.230 km

With displacement calculated, we can find the average speed as follows:

V = S/t  ⇒  V = \frac{2.230}{1.750}

V = 1.274\frac{km}{h}

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3 years ago
The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
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Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

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3 years ago
A rock displaces 1.65 L of water. The volume of the rock is:
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Answer:

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Hence volume of rock is = 1.65L or 1650 cm^3

Explanation:

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