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ryzh [129]
3 years ago
6

What is meant by the term fitness​

Physics
1 answer:
Artemon [7]3 years ago
4 0

Answer:

An organisms ability to survive and reproduce in a particular environment.

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Mercury is 0.39 AU from the sun. What is its distance from the sun in kilometers?
qaws [65]

Answer:

0.39AU = 58344000 [km] or 58.500.000 [km]

Explanation:

This is a problem that consists simply in converting units:

AU = astronomic unit

1 [AU] = 1.496*10^{8} [km]\\0.39 [AU] = x\\\\Therefore\\x = \frac{0.39*1.496*10^{8} }{1} \\x = 58344000[km]

8 0
3 years ago
If a gas produced, most likely a ______ reaction took place ​
Orlov [11]

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if a gas produced, most likely a chemical reaction took place ​

Explanation:

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3 years ago
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A well labelled diagram of outdoor thermometer​
ElenaW [278]
Hope it helps u! :)

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3 years ago
Importance of SI system in points​
monitta

Answer:

SI unit is an international system of measurements that are used universally in technical and scientific research to avoid the confusion with the units. Having a standard unit system is important because it helps the entire world to understand the measurements in one set of unit system.

4 0
3 years ago
A 2000 kg car moves along a horizontal road at speed vo
cluponka [151]

Answer:

The shortest possible stopping distance of the car is 175.319 meters.

Explanation:

In this case we see that driver use the brakes to stop the car by means of kinetic friction force. Deceleration of the car is directly proportional to kinetic friction coefficient and can be determined by Second Newton's Law:

\Sigma F_{x} = -\mu_{k}\cdot N = m \cdot a (Eq. 1)

\Sigma F_{y} = N-m\cdot g = 0 (Eq. 2)

After quick handling, we get that deceleration experimented by the car is equal to:

a = -\mu_{k}\cdot g (Eq. 3)

Where:

a - Deceleration of the car, measured in meters per square second.

\mu_{k} - Kinetic coefficient of friction, dimensionless.

g - Gravitational acceleration, measured in meters per square second.

If we know that \mu_{k} = 0.0735 and g = 9.807\,\frac{m}{s^{2}}, then deceleration of the car is:

a = -(0.0735)\cdot (9.807\,\frac{m}{s^{2}} )

a = -0.721\,\frac{m}{s^{2}}

The stopping distance of the car (\Delta s), measured in meters, is determined from the following kinematic expression:

\Delta s = \frac{v^{2}-v_{o}^{2}}{2\cdot a} (Eq. 4)

Where:

v_{o} - Initial speed of the car, measured in meters per second.

v - Final speed of the car, measured in meters per second.

If we know that v_{o} = 15.9\,\frac{m}{s}, v = 0\,\frac{m}{s} and a = -0.721\,\frac{m}{s^{2}}, stopping distance of the car is:

\Delta s = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(15.9\,\frac{m}{s} \right)^{2}}{2\cdot \left(-0.721\,\frac{m}{s^{2}} \right)}

\Delta s = 175.319\,m

The shortest possible stopping distance of the car is 175.319 meters.

8 0
4 years ago
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