a 40 kg rock lies at a bottom of a lake. The volume of water it displaces is 800cm³. Calculate the minimum force required to lift the rock from the bottom of the lake.​

Answer:

what question is that?

Explanation:

make it english

Except when acted upon by an imbalanced force, an object at rest remains at rest and an object in motion continues to move in a straight path at a constant pace.

According to the first law, a force must be applied to an object before it may affect its motion. According to the second law, an object experiences a force equal to its mass times its acceleration. According to the third law, when two things interact, they exert forces on one another that are both equal in strength and directed in the opposite direction.

According to the first law, an object will continue moving in the same direction unless another force acts on it. According to the second law, an object's force is determined by multiplying its mass by its acceleration. There is an equal and opposite reaction to every action, according to the third law.

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Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

105 kg*m/s

Explanation:

p=mv

momentum=mass*velocity

momentum=15kg*7m/s

momentum=105kg*m/s

In any given week, I think I am exposed to sounds loud enough to cause hearing damage about 2-3 times. I am not using hearing protection at those times. I do not think I have already experienced hearing loss, but I am not sure.

Hearing loss refers to a partial or total inability to hear sounds in one or both ears. It can vary in severity, from mild to profound, and can affect people of all ages.

I think there are a few reasons why people don't use hearing protection more often. One reason is that they may not be aware of the risks of noise-induced hearing loss. Another reason is that they may not think the risks are high enough to warrant using hearing protection. Finally, some people may find hearing protection uncomfortable or inconvenient to use.

Here are some things that could be done to encourage people to protect their hearing:

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Three facts I have learned this year in Physical Education are:

a) The importance of stretching before and after exercising.

b) The benefits of cardiovascular exercise.

c) The importance of staying hydrated during exercise.

Firstly, I have learned the importance of stretching before and after exercising. Stretching helps prevent injury and improves flexibility. I will make sure to incorporate stretching into my daily routine, whether it's through a short stretching session in the morning or by taking a few minutes to stretch before and after a workout.
Secondly, I have learned the benefits of cardiovascular exercise. Cardiovascular exercise, such as running or cycling, helps improve heart health and endurance. I will try to incorporate more cardiovascular exercise into my daily routine, such as going for a run or bike ride after work.
Lastly, I have learned the importance of staying hydrated during exercise. Dehydration can lead to fatigue and muscle cramps, so it's important to drink plenty of water before, during, and after exercise. I will make sure to carry a water bottle with me throughout the day and refill it regularly to ensure that I am staying hydrated.
In summary, by incorporating stretching, cardiovascular exercise, and staying hydrated into my daily routine, I hope to improve my overall physical health and well-being.

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Answer:

Coulomb's law calculates the magnitude of the force F between two point charges, q1 and q2, separated by a distance r. F=k|q1q2|r2.

Explanation:

Answer:

B It is always increasing.

Explanation:

In Physics, entropy can be defined as the tendency or ability of a substance to reach maximum disorder i.e to be randomly distributed.

This ultimately implies that, entropy is a thermodynamic quantity that measures the degree of maximum disorder or randomness of a system.

The S.I unit used for the measurement of the degree of maximum order or randomness of a system is Joules per Kelvin (JK¯¹). An example of entropy is the mixing of ideal gases.

Generally, the entropy in an irreversible process always increases and as such the change in entropy has a positive value.

Hence, the entropy of the universe is always increasing because its energy flow is considered to be in a downward direction rather than upward i.e from a hot region to a cold region; making the energy to be evenly distributed.

Answer:

14.24 N

Explanation:

First, we need to calculate the mass of the dog. It can be calculated as the weight on Earth divided by the gravity on Earth, so

m = 178 N/ 9.8 m/s²

m = 18.16 kg

Then, the weight of the dog on pluto is equal to the mass times the gravity on pluto, so

W = (18.16 kg)(0.784 m/s²)

W = 14.24 N

Therefore, the answer is

14.24 N

Answer:

1,840,000 J

Explanation:

The energy required for a particular change in state is given by the specific latent heat. Specific latent heat is the amount of energy required to change the state of 1 ... of ice into 1 kg of water at its melting point of 0°C. The same amount of energy ... stored or released as the temperature of a system changes can be calculated.

X - rays are highly penetrating electromagnetic radiations that are formed when the cathode rays strikes a dense metal.

It has many characteristics such as :

Answer:

The answer is below

Explanation:

a) Using the formula:

\(\omega^2=\omega_o^2+2\alpha \theta\\\\\omega=final\ angular\ velocity,\omega_o=initial\ anglular\ velocity,\alpha= angular\ acceleration,\\\theta=angular\ distance\\\\Given\ that:\\\\initial\ velocity(u)=28.78m/s,distance(s)=50\ m,radius(r)=0.25\ m,\\final/ velocity(v)=0(stop)\\\\\omega=v/r=\frac{28.78m/s}{0.25m} =115.12\ rad/s,\omega_o=0,\theta=s/r=\frac{50\ m}{0.25\ m}=200\ rad\\ \\\omega^2=\omega_o^2+2\alpha \theta\\\\115.12^2=0^2+2\alpha(200)\\\\2\alpha(200)=13252.6144\\\\\alpha=33.13\ rad/s^2\)

b)

\(\theta=200\ rad=200\ rad*\frac{1\ rev}{2\pi\ rad}=31.83\ rev\)

When the pressure (100 N) and weight (40 kg) are entered into the appropriate fields, the gravity is 2 m/s².

Force and acceleration are linked by the equation F=ma. The characters "F," "m," and "a" stand for acceleration, mass, and force, respectively. Force is the ability of one object to exert a pull or force on another. Acceleration is the rate at which an object's speed changes.

Speed increase is the pace of progress of speed. Acceleration typically indicates a change in speed, but not always. Because its velocity is changing in the opposite direction, even if an object moves in a circle at the same speed, it will still accelerate.

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Fundamental frequency :

f = (1/2L) √(T/p)

Where:

L = lenght

T = tension

p = mass

Frequency is inversely proportional to Lenght

Frequency is proportional to the square root of the tension

If both are doubled:

f'= f * (1/2) √2

f' = f / √2 = 2048 / √2 = 1448 HZ

The contact force (in N) between each pair of boxes is mathematically given as

F_{ab} = 3.56 N

F_{bc} = 1.2 N

Generally, Any force that is generated as a consequence of two objects coming into touch with one another is referred to as a contact force. Contact forces are present everywhere and are the cause of the vast majority of macroscopic groupings of matter's obvious interactions with one another.

In conclusion, The equation for is  Acceleration of the system is mathematically given as

a = 8.25 / (5.85 + 2.95 + 1.50)

a= 0.8 m/s^2

Therefore

F_{ab} = (Mb + Mc)*a

F_{ab} = (2.95 + 1.50) * 0.8

F_{ab} = 3.56 N

F_{bc} = Mc * a

F_{bc}= 1.5 * 0.8

F_{bc} = 1.2 N

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After considering the given data we come to the conclusion that the increase in the internal energy of the tie and spike is 16.322 J, under the condition that  a given weight of  0.76 kg spike had been jamed into a railroad tie .

In order to evaluate the increase in internal energy of the tie and spike we have to apply the formula of kinetic energy which is
1/2 × mv²
Here,
m = mass
v = velocity
Staging the values in the formula
Kinetic energy of the spike = (1/2)× (0.76)×(4.8)²= 87.2064 J
Energy absorbed by the tie and spike = 87.2064 J × 0.187 = 16.322 J

Hence, the evaluate rise in internal energy of the tie and spike is 16.322 J.
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An action which would help in preventing and coping with heat-related conditions is: A. Drinking water.

Heat can be defined as a form of energy that is transferred from a physical object (body) to another, as a result of a difference in temperature. Also, heat is a condition of weather that is generally characterized by a high degree of temperature.

This ultimately implies that, heat is most likely to cause dehydration and high body temperature.

In order to prevent and cope with heat-related conditions, you should ensure that you drink water at regular intervals for hydration.

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One of the things that are NOT a typical consequence that is associated with open pit mines is the release of high concentrations of ozone into the local environment, causing long-term respiratory harm to individuals and the environment.

The open-pit mine is a surface mining technique used to extract rocks or minerals from the earth using an open-air pit. This technique is used when ore or rock deposits are found near the surface, which makes digging a tunnel unnecessary.

The main consequence of open-pit mine that concerns most people is that it generates a large amount of dust that contains heavy metals to the environment, as well as creating deposits of toxic waste ponds. Ozone is not considered a heavy metal.

Attached below is an image of the Sunrise Dam Gold Mine open pit mine in Australia.

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(a) The magnitude of the wheels angular acceleration is 0.0088 rad/s².

(b) The magnitude of the tangential acceleration after the maximum operational speed is reached is 0.153 m/s².

The angular acceleration of the wheel is calculated as follows;

α = ω/t

ω = v/r

α = v/(rt)

α = (2.3)/(17.5 x 15)

α = 0.0088 rad/s²

a = v/t

a = (2.3)/15

a = 0.153 m/s²

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Answer:

Its nymber 2

Explanation:

Answer:

5m

Explanation:

(500 joules) / (100 newtons) =

5 meters

The magnetic force per unit length of the wire is (C) (0.49i - 0.51j - 1.37k) N/m.

To calculate the magnetic force per unit length of the wire, we can use the formula:

F = I * (L x B),

where F is the force, I is the current, L is the length vector of the wire, and B is the magnetic field.

Given:

Current, I = 3A

Length vector, L = √√3 * (i + j + k)

Magnetic field, B = 0.75i + 0.4k

Let's calculate the cross product of L and B:

L x B = | i j k |

|√√3 √√3 √√3|

|0.75 0 0.4|

To evaluate this cross product, we calculate the determinants:

(i) component: (√√3 * 0 - √√3 * 0.4) = -0.4√√3

(j) component: (-√√3 * 0.75 - √√3 * 0) = -0.75√√3

(k) component: (√√3 * 0.75 - √√3 * 0) = 0.75√√3

Now, multiply the cross product by the current:

F = 3A * (-0.4√√3i - 0.75√√3j + 0.75√√3k)

Simplifying this expression gives:

F = (-1.2√√3i - 2.25√√3j + 2.25√√3k) N

Therefore, the magnetic force per unit length of the wire is approximately (-1.2√√3i - 2.25√√3j + 2.25√√3k) N/m.

Comparing the given answer options, the closest match is C. (0.49i - 0.51j - 1.37k) N/m.

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The minimum velocity for the meteorite before entrance into earth's atmosphere is 36.8 m/s.

To calculate the minimum velocity the meteorite must have had before it entered Earth's atmosphere, the equation for heat transfer is:

Q = mcΔT

Where Q = the heat transferred,

m = the mass of the meteorite,

c = the specific heat capacity of iron, and

ΔT = the change in temperature.

The specific heat capacity of iron is approximately = 0.45 J/g°C.

Temperature of the meteorite changes from -105°C to the melting point of iron, which is approximately = 1535°C.

So: Q = mc(1535 - (-105))

Mass of the meteorite unknown, but we know it is made of iron, so assume a value for the mass, such as 100 g.

Q = 100 x 0.45(1535 - (-105))

Solving for Q:

Q = 67875 J

To calculate the kinetic energy of the meteorite, which is given by the equation: Ek = 1/2mv²

where Ek is kinetic energy,

m = the mass of the meteorite and

v = the velocity of the meteorite.

Therefore, the velocity of the meteorite is given by the following equation: v = √(2Ek/m)

Substituting the values of Ek and m, we get

v = √(2 x 67875/100)

Solving for v:

v = √1357.5

The minimum velocity the meteorite must have had before it entered Earth's atmosphere is approximately 36.8 m/s.

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Answer:

Examples include cars, buildings, clothing, and tools. Nonmaterial culture refers to the abstract ideas and ways of thinking that make up a culture. Examples of nonmaterial culture include traffic laws, words, and dress codes.

Answer:

2.4 m/s

Explanation:

Applying,

The law of conservation of momentum

Total momentum before collision = Total momentum after collision

mu+m'u' = mv+m'v'.................... Equation 1

Where m = mass of the moving puck, m' = mass of the stationary puck, u = initial velocity of the moving puck, u' = initial velocity of the stationary puck, v = final velocity of the moving puck, v' = final velocity of the stationary puck.

make v' the subject of the equation

v' = (mu+m'u'-mv)/m'................ Equation 2

From the question,

Given; m = 3 kg, m' = 2 kg, u = 2 m/s, u' = 0 m/s (stationary), v = 0.4 m/s

Substitute these values into equation 2

v' = [(3×2)+(2×0)-(3×0.4)]/2

v' = (6-1.2)/2

v' = 4.8/2

v' = 2.4 m/s

(a) The mass of water is approximately 49.65 kg. (b) The final temperature of the water vapor will be 120°C. (c) The total enthalpy change is approximately 277,956 kJ. (d) Diagram shown below.

(a) To determine the mass of the water, we need to know its density at the given conditions. The density of water changes with temperature and pressure. At 40°C and 200 kPa, the density of water is approximately 993 kg/m³.

Since we have 50 L of water, we need to convert it to cubic meters:

50 L = 0.05 m³

Now we can calculate the mass of water:

Mass = Density * Volume

Mass = 993 kg/m³ * 0.05 m³

Mass ≈ 49.65 kg

Therefore, the mass of water is approximately 49.65 kg.

(b) To find the final temperature, we need to consider the phase change from liquid to vapor. At constant pressure, the temperature will remain constant until all the liquid water has vaporized. This temperature is called the saturation temperature.

We can determine the saturation temperature at 200 kPa using a steam table or other relevant data sources. Let's assume that the saturation temperature is 120°C.

Therefore, the final temperature of the water vapor will be 120°C.

(c) To calculate the total enthalpy change, we need to consider the energy required to heat the water from its initial temperature to the saturation temperature, as well as the energy required for the phase change from liquid to vapor.

The enthalpy change during heating can be calculated using the formula:

ΔH1 = Mass * Specific Heat Capacity * ΔT1

Where:

Mass = 49.65 kg (from part a)

Specific Heat Capacity = specific heat capacity of water at constant pressure = 4.18 kJ/(kg·°C)

ΔT1 = final temperature - initial temperature = 120°C - 40°C = 80°C

ΔH1 = 49.65 kg * 4.18 kJ/(kg·°C) * 80°C

ΔH1 ≈ 165,938 kJ

The enthalpy change during phase change can be calculated using the formula:

ΔH2 = Mass * Latent Heat of Vaporization

Where:

Mass = 49.65 kg (from part a)

Latent Heat of Vaporization = energy required to vaporize 1 kg of water = 2257 kJ/kg

ΔH2 = 49.65 kg * 2257 kJ/kg

ΔH2 ≈ 112,018 kJ

The total enthalpy change is the sum of ΔH1 and ΔH2:

Total Enthalpy Change = ΔH1 + ΔH2

Total Enthalpy Change ≈ 165,938 kJ + 112,018 kJ

Total Enthalpy Change ≈ 277,956 kJ

Therefore, the total enthalpy change is approximately 277,956 kJ.

(d) The process can be shown on a T-v (temperature-volume) diagram with respect to saturation lines. In this case, the process starts at the initial temperature and pressure (40°C, 200 kPa), and moves along the constant pressure line until reaching the saturation temperature (120°C). Then, the process follows the saturation line until the entire liquid is vaporized.

Here is a simplified representation of the process on a T-v diagram:

           |

Saturation |                       |

 Line   |                             |

           |                             |

           |                             |

           |                             |

           |                             |

           |                             |

           |                             |

 Initial |-----------------------------| Final

           |                             |

           |                             |

           |                             |

           |                             |

           |                             |

           |                            

This diagram is a rough representation and does not accurately reflect specific volume values or scale. It simply illustrates the general process from initial conditions to the final state along the constant pressure and saturation lines.

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Answer:

yez

Explanation:

ANSWER:

False

STEP-BY-STEP EXPLANATION:

Even without friction, the efficiency cannot be greater than 100%, because the output cannot be greater than the input, therefore, the statement is false.