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I am Lyosha [343]
3 years ago
7

Determine the gravitational force of attraction experienced by a two 5.00 kilogram masses separated by a distance of 2.5 meters.

Physics
1 answer:
drek231 [11]3 years ago
4 0
The gravitational force between two objects may be calculated through the equation,
                                     F = Gm₁m₂ / d²
where G is 6.67 x 10^-11 m³/s²kg
Substituting the known values to the equation,
                        F = (6.67 x 10^-11 m³/s²kg)(5.0 kg)(5.0 kg) / (2.5 m)²
                                     F = 2.668 x 10^-10 N
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In a nuclear power plant, the temperature of the water in the reactor is above 100°C because of what?
SVETLANKA909090 [29]

Answer:

The temperature of the water increases because the nuclear reactor heats it producing steam

Explanation:

The nuclear power plants are usually defined as those thermal plants where the nuclear reactors are used in order to generate heat that eventually leads to the rotating of the turbines and produces electricity. Here the nuclear reactor heats the water, and it increases above a temperature of 100°C, where this heat energy plays a key role in the entire process. It is an efficient method as it does not lead to the emission of any green house gases that are harmful to the environment.

8 0
3 years ago
You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=
velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

Hence, the time it takes to achieve a (vertical) velocity change of \Delta v_{y} would be:

\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

8 0
2 years ago
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taurus [48]

Answer:

4m

Explanation:

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3 years ago
What is the dependent variable in an experiment that shows how the volume of a gas changes with change in temperature?
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