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3 years ago

Birds, squirrels, and chipmunks all living in the same tree, using the same resources, are an example of:

Physics

1 answer:

Ulleksa [173]3 years ago

4 0

Explanation :

In Interspecific competition, different individuals of different species fight for the same resource in an ecosystem. The resources can be food or living space.

Birds belong to the group of aves, squirrels and chipmunks are from the family of Sciuridae.

All three are from different species. They are on the same ecosystem i.e. tree. They form the interspecific competitions.

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Help me with Economics please and thank you the question is going to be down in a attached file while the answers on here

Tresset [83]

I think is A or B it depends on like what the trying to answer

6 0

3 years ago

Read 2 more answers

An airplane during departure has a constant acceleration of 3 m / s².

Rama09 [41]

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

Acceleration = speed/time

3m/s² = speed/4s

S = 12m/s

The speed of the plane after 4s is 12m/s.

b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = v - u

3 = v - 20

24 = v - 20

v = 44m/s

After 8s the plane would've 44m/s speed.

6 0

3 years ago

Suppose that the Mars orbiter was to have established orbit at 155 km and that one group of engineers specified this distance as

MAXImum [283]

Answer:

108 km

Explanation:

The conversion factor between meters and feet is

1 m = 3.28 ft

So the second altitude, written in feet, can be rewritten in meters as

$h_2 = 1.55 \cdot 10^5 ft \cdot \frac{1}{3.28 ft/m}=4.7\cdot 10^4 m$

or in kilometers,

$h_2 = 47 km$

the first altitude in kilometers is

$h_1 = 155 km$

so the difference between the two altitudes is

$\Delta h = 155 km - 47 km = 108 km$

8 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

A supersonic nozzle is also a convergent–divergent duct, which is fed by a large reservoir at the inlet to the nozzle. In the re

Lady_Fox [76]

Answer:

155.38424 K

2.2721 kg/m³

Explanation:

$P_1$ = Pressure at reservoir = 10 atm

$T_1$ = Temperature at reservoir = 300 K

$P_2$ = Pressure at exit = 1 atm

$T_2$ = Temperature at exit

$R_s$ = Mass-specific gas constant = 287 J/kgK

$\gamma$ = Specific heat ratio = 1.4 for air

For isentropic flow

$\frac{T_2}{T_1}=\frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=T_1\times \frac{P_2}{P_1}^{\frac{\gamma-1}{\gamma}}\\\Rightarrow T_2=00\times \left(\frac{1}{10}\right)^{\frac{1.4-1}{1.4}}\\\Rightarrow T_2=155.38424\ K$

The temperature of the flow at the exit is 155.38424 K

From the ideal equation density is given by

$\rho_2=\frac{P_2}{R_sT_2}\\\Rightarrow \rho=\frac{1\times 101325}{287\times 155.38424}\\\Rightarrow \rho=2.2721\ kg/m^3$

The density of the flow at the exit is 2.2721 kg/m³

4 0

3 years ago