A football thrown by a professional quarterback goes farther than one thrown by a 10-year old. What Newton Law is this?

Answer:

it goes up until we help it to but the moment we stop support it gets affected by gravity and goes back

Explanation:

We learn about objects of interest to intelligence through matter/energy interactions such as emission, reflection, refraction, and absorption.

Emission: Objects can emit energy in the form of light, heat, or other types of radiation. By detecting and analyzing the emitted radiation, we can gather information about the object's properties and composition.
Reflection: When light or other forms of energy bounce off an object's surface, we can observe and analyze the reflected radiation. The characteristics of the reflected radiation can provide insights into the object's shape, color, and surface properties.
Refraction: When energy passes through a medium and changes direction, such as when light bends while passing through a transparent object, it undergoes refraction. By studying the changes in the direction and intensity of the refracted energy, we can gain knowledge about the object's composition and structure.
Absorption: Objects can absorb certain types of energy, causing a decrease in its intensity. By examining the absorbed energy and the wavelengths that are absorbed, we can acquire information about the object's chemical composition and properties.
Through these interactions, scientists and researchers employ various instruments and techniques to gather data and learn about objects of interest, enabling us to deepen our understanding and make informed interpretations and analyses.

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Int he abve image relating to rock cycle, A = Igneous Rock

B = Metamorphic Rock

C = Sedimentary Rock.

The rock cycle is a continuous process that describes the transformation of rocks through various geological processes. It involves the formation, breakdown, and reformation of three main types of rocks

The cycle starts with the formation of igneous rocks through the solidification of molten magma or lava. These rocks can then be weathered and eroded into sediments,which are compacted and cemented to form   sedimentary rocks.

Under intense heat and pressure,these rocks can undergo metamorphism, resulting in   the formation of metamorphic rocks.

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As you move from the positive plate to the negative plate in the gap of a parallel plate capacitor, the electric field is directed from the positive plate to the negative plate. The electric field lines are parallel and uniform between the plates.

This is because the positive plate creates a positive electric field pointing away from it, while the negative plate creates a negative electric field pointing towards it. The net result is a uniform electric field directed from positive to negative.

The magnitude of the electric field remains constant throughout the gap between the plates, assuming there are no external influences or variations. This uniform electric field distribution is a characteristic of a parallel plate capacitor and is essential for its functioning in storing electric charge.

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Answer:

74.88N

Explanation:

From the question,

F = ma................... Equation 1

Where F = force exerted on the ball, m = mass of the ball, a = acceleration

But,

v² = u²+2as.............. Equation 2

Where v = final velocity, u = initial velocity, s = distance.

Given: v = 12 m/s, u = 0 m/s (from rest), s = 5.0 m

Substitute into equation 2 and solve for a

12² = 0²+2×a×5

144 = 10a

10a = 144

a = 144/10

a = 14.4 m/s²

Also Given: m = 5.2 kg,

Substitute into equation 1

F = 5.2×14.4

F = 74.88 N

Hence the force exerted on the ball is 74.88 N

The energy required to increase the temperature is 5277.78 J

Here we want to find the heat energy required to raise the temperature of a substance, so we can use the formula:

Q = m * c * ΔT

Where:

In your case, the values are:

m = 37.5 g (mass of water)

c = 4.184 J/g°C (specific heat capacity of water)

ΔT = (55.2°C - 23.0°C) = 32.2°C (change in temperature)

Now, let's substitute these values into the formula:

Q = 37.5 g * 4.184 J/g°C * 32.2°C

Q = 5277.78 J

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Kinetic energy is the energy possessed by a body as a result of its motion. Therefore, the kinetic energy of a 625-kg roller coaster car that is moving with a speed of 18.3 m/s is 104,640.625 J which is closest to option A, i.e., 127,575 J.

It is calculated by multiplying half of the mass of a body with the square of its velocity. The kinetic energy formula can be written as, KE = (1/2)mv2Where,KE is the kinetic energy of the body, m is the mass of the body, v is the velocity of the body. Now, let us apply the above formula to find the kinetic energy of the given roller coaster car whose mass is 625 kg and speed is 18.3 m/s.KE = (1/2)mv2KE = (1/2) x 625 x (18.3)2KE = (1/2) x 625 x 334.89KE = 104,640.625 J.

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Answer: direction

Explanation:

Given

The resultant vector of a force gives us information regarding the direction of the resultant force.

If there are multiple forces acted in a different direction then, the resultant vector describes the direction of the resultant force.

The turbine generates entropy at a rate of about 2.4944 kW/K. The option that comes closest to the provided values is (c) 3.34 kW/K.

To find the rate of entropy generation in the turbine, we need to apply the concept of entropy balance. The rate of entropy generation can be determined by calculating the difference between the entropy flow into and out of the system.

Given:

Inlet conditions:

Pressure at inlet (P₁) = 0.18 MPa

Mass flow rate (m) = 1.6 kg/s

Exit conditions:

Pressure at exit (P₂) = 1 MPa

Temperature at exit (T₂) = 60 degrees C = 333.15 K

First, we need to determine the specific entropy at the inlet and outlet states. We can use the properties of Refrigerant-134a to find these values.

From the saturation table for Refrigerant-134a at 0.18 MPa (inlet pressure), we can find the corresponding saturation temperature T1.

At P₁ = 0.18 MPa:

Saturation temperature T1 = 20.83 degrees C = 293.98 K

From the superheated table for Refrigerant-134a at 1 MPa (exit pressure) and 60 degrees C (exit temperature), we can find the specific entropy value S2.

At P₂ = 1 MPa, T₂ = 60 degrees C:

Specific entropy S₂ = 1.559 kJ/(kg·K)

The rate of entropy generation in the turbine can be calculated as:

Rate of entropy generation = m * (S₂ - S₁)

Where:

m = mass flow rate

S₂ = Specific entropy at the exit

S₁ = Specific entropy at the inlet

Substituting the values:

Rate of entropy generation = 1.6 kg/s * (1.559 kJ/(kg·K) - 0)

Rate of entropy generation = 1.6 kg/s * 1.559 kJ/(kg·K)

Rate of entropy generation ≈ 2.4944 kW/K

Therefore, the rate of entropy generation in the turbine is approximately 2.4944 kW/K.

Among the given options, the closest one is (c) 3.34 kW/K.

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Answer:

19.6 cm.

Explanation:

From the question given above, the following data were obtained:

Focal length (f) = 13.8 cm

Magnification (M) = +2.37

Object distance (u) =.?

Next, we shall determine the image distance. This can be obtained as follow:

Magnification (M) = +2.37

Object distance (u) = u

Image distance (v) =?

M = v / u

2.37 = v / u

Cross multiply

v = 2.37 × u

v = 2.37u

Finally, we shall determine the object distance. This can be obtained as follow:

Focal length (f) = 13.8 cm

Image distance (v) = 2.37u

Object distance (u) =.?

1/v + 1/u = 1/f

vu / v + u = f

2.37u × u / 2.37u + u = 13.8

2.37u² / 3.37u = 13.8

Cross multiply

2.37u² = 3.37u × 13.8

2.37u² = 46.506u

Divide both side by u

2.37u² / u = 46.506u / u

2.37u = 46.506

Divide both side by 2.37

u = 46.506 / 2.37

u = 19.6 cm

Thus, the lens should be held at a distance of 19.6 cm.

(a) The initial rate of change of the plate temperature is -0.163 K/s.

(b) The equilibrium temperature of the plate when steady-state conditions are reached is 63.5°C.

(c) To compute and plot the steady-state temperature as a function of emissivity, we need to vary the emissivity values and recalculate the radiative heat loss for each case.

(a) Initial Rate of Change of Plate Temperature:

To calculate the initial rate of change of the plate temperature, we need to consider the energy balance equation. The equation is given by:

ρcA(dT/dt) = Q_in - Q_out

Where:

ρ is the density of aluminum (2700 kg/m³)

c is the specific heat of aluminum (900 J/kg · K)

A is the surface area of the plate

(dT/dt) is the rate of change of temperature

Q_in is the solar radiation absorbed

Q_out is the heat loss through convection

First, let's calculate the surface area of the plate:

Given thickness of the plate = 4 mm = 0.004 m

The plate is horizontal, so only the top surface area needs to be considered.

Assuming the plate has a square shape, let's say its length and width are L.

The surface area is then A = L * L = L²

Given:

Solar radiation incident flux, Q_in = 900 W/m²

Absorption coefficient of the coating, α = 0.8

Emissivity of the coating, ε = 0.25

Convection heat transfer coefficient, h = 20 W/m² · K

Now, let's calculate the initial rate of change of temperature:

ρcA(dT/dt) = αQ_in - εσA(T⁴ - T_a⁴) - hA(T - T_a)

Where:

σ is the Stefan-Boltzmann constant (σ ≈ 5.67 × 10⁻⁸ W/m² · K⁴)

T is the temperature of the plate (initially unknown)

T_a is the ambient air temperature

Rearranging the equation, we get:

ρc(dT/dt) = αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)

Now, we have all the required values to solve this equation.

(b) Equilibrium Temperature:

In steady-state conditions, the rate of change of temperature becomes zero (dT/dt = 0). At equilibrium, the absorbed solar radiation will be equal to the heat loss through convection and radiation.

αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)

We need to solve this equation to find the equilibrium temperature, T_eq.

(c) Variation of Steady-State Temperature with Emissivity:

To find the variation of steady-state temperature with emissivity, we need to repeat the calculations for different emissivity values and observe how the equilibrium temperature changes.

Let's start by solving part (a):

(a) Initial Rate of Change of Plate Temperature:

Using the equation:

ρc(dT/dt) = αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)

Substituting the given values:

ρ = 2700 kg/m³

c = 900 J/kg · K

α = 0.8

Q_in = 900 W/m²

ε = 0.25

σ = 5.67 × 10⁻⁸ W/m² · K⁴

T_a = ambient air temperature (not provided)

h = 20 W/m² · K

A = L² (surface area, to be determined)

We can simplify the equation by dividing both sides by ρc:

(dT/dt) = [αQ_in - εσ(T⁴ - T_a⁴) - h(T - T_a)] / (ρc)

Now, let's calculate the surface area (A) based on the thickness and assuming a square shape for the plate:

Given:

Thickness of the plate, t = 4 mm = 0.004 m

Area of the top surface = A

A = L²

Since the plate is square-shaped, L = √(A).

Now, we can substitute the values and solve for (dT/dt):

(dT/dt) = [0.8 * 900 - 0.25 * (5.67 × 10⁻⁸) * (T⁴ - T_a⁴) - 20 * (T - T_a)] / (2700 * 900)

This gives us the initial rate of change of the plate temperature.

(b) Equilibrium Temperature:

Using the equation:

αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)

We can rearrange the equation to solve for the equilibrium temperature (T_eq):

αQ_in = εσA(T⁴ - T_a⁴) + hA(T - T_a)

0.8 * 900 = 0.25 * (5.67 × 10⁻⁸) * A * (T_eq⁴ - T_a⁴) + 20 * A * (T_eq - T_a)

Simplifying further:

720 = 0.25 * (5.67 × 10⁻⁸) * A * (T_eq⁴ - T_a⁴) + 20 * A * (T_eq - T_a)

Now, we can solve this equation to find the equilibrium temperature (T_eq).

(c) Variation of Steady-State Temperature with Emissivity:

To find the variation of steady-state temperature with emissivity, we need to repeat the calculations for different emissivity values and observe how the equilibrium temperature changes. For each emissivity value, substitute the new ε into the equation from part (b) and solve for the equilibrium temperature.

Repeat the calculations for ε = 0.1, 0.5, and 1, and observe the variations in equilibrium temperature. Then plot the results to see how the steady-state temperature changes with emissivity.

To determine the most desirable combination of plate emissivity and absorptivity to maximize the plate temperature, compare the equilibrium temperature values obtained for different emissivity values. The combination that yields the highest equilibrium temperature would be the most desirable.

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The radii r₁, r₂, and r₃ of the nuclei 4,2He, 236,92U, and 56,26Fe are  1.9044 x 10⁻¹⁵ meters, 4.3944 x 10⁻⁵ meters, 4.3944 x 10⁻¹⁵ meters.

The radii of atomic nuclei can be estimated using the empirical formula known as the "constant density model." According to this model, the radius (r) of a nucleus can be approximated using the equation:

r = r0 A¹/³

where r0 is a constant and A is the mass number of the nucleus.

The value of r0 is typically taken to be around 1.2 fm (femtometers) or 1.2 x 10⁻¹⁵ meters.

For the nucleus 4,2He (helium-4):

A = 4

r0 = 1.2 fm

r1 = 1.2 fm × 4¹/³

≈ 1.2 fm × 1.587

≈ 1.9044 fm

≈ 1.9044 x 10⁻¹⁵ meters

Therefore, r1 = 1.9044 x 10⁻¹⁵ meters.

For the nucleus 236,92U (uranium-236):

A = 236

r0 = 1.2 fm

r2 = 1.2 fm × 236¹/³

≈ 1.2 fm × 6.118

≈ 7.3416 fm

≈ 7.3416 x 10⁻¹⁵ meters

Therefore, r2 = 7.3416 x 10⁻¹⁵ meters.

For the nucleus 56,26Fe (iron-56):

A = 56

r0 = 1.2 fm

r3 = 1.2 fm × 56¹/³

≈ 1.2 fm × 3.662

≈ 4.3944 fm

≈ 4.3944 x 10⁻¹⁵ meters

Therefore, r3 = 4.3944 x 10⁻¹⁵ meters.

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Alpha Centauri, the star closest to the sun, is located 4.3 light years away. Alpha Centauri and Earth are separated by around 4.1 × 10¹⁶ meters.

To convert the distance of 4.3 light-years (ly) to meters, we can use the conversion factor of 1 light-year equal to 9.461 × 10¹⁵ meters. Multiplying 4.3 by this conversion factor gives us the distance in meters:

4.3 ly * 9.461 × 10¹⁵ meters/ly = 4.0853 × 10¹⁶ meters

Rounding to two significant figures, the distance to Alpha Centauri is approximately 4.1 × 10¹⁶ meters. This distance represents the vast scale of interstellar distances.

Alpha Centauri is the closest star system to our solar system, yet its distance is still incredibly immense. Understanding these astronomical distances helps us appreciate the vastness of the universe and the challenges involved in space exploration and interstellar travel.

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The energy used for the 30-day period is 1,355,000 kJ, the average energy used is 45,166.67 kJ/day, and the electric bill for this month would be approximately $101.41.

To find the energy used for the 30-day period in kJ: a) Energy Used (kJ) = 1355 kWh * 1000 Wh/kWh = 1,355,000 Wh = 1,355,000 kJ

To find the average energy used in J/day: b) Average Energy Used (J/day) = Energy Used (kJ) / 30 days

Average Energy Used (J/day) = 1,355,000 kJ / 30 days = 45,166.67 kJ/day

To calculate the electric bill for this month: c) Electric Bill = Energy Used (kWh) * Cost per kWh Electric Bill = 1355 kWh * $0.0749/kWh = $101.4145. Therefore, your electric bill for this month would be approximately $101.41.

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The value of resistance R must be 7.77 x 10⁴ Ω to give a percent decrease in angular frequency of 5.0% compared to the value calculated in part A.

In part A,

the angular frequency is ω = 3.3 x 10⁵ rad/s.

To find the value of resistance R to give a 5.0% decrease in angular frequency, the following formula is used,

ω' = ω (1 - δ)

where

ω is the original angular frequency,

ω' is the new angular frequency,

δ is the percentage decrease.

Substituting the given values,

ω' = 3.3 x 10⁵ rad/s (1 - 5.0/100)

ω' = 3.135 x 10⁵ rad/s

Now we can use the formula for angular frequency to calculate the value of resistance R as follows:

ω = 1/√(LC - R²)

R = √(LC - ω'²)

R = √((0.420 H)(2.50 x 10⁻⁵ F) - (3.135 x 10⁵ rad/s)²)

R = 7.77 x 10⁴ Ω

Therefore, the value of resistance R must be 7.77 x 10⁴ Ω to give a percent decrease in angular frequency of 5.0% compared to the value calculated in part A.

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18.22 kg/mol is the molar mass of the gene fragment if a bacterial gene fragment of 10.0 mg is dissolved in enough water to make 30.0 ml of solution.

What is a solution?

In a homogenous mixture of two or more components, a solution is defined as having particles less than one nanometer in size. Solutions come in many forms, such as sugar and salt solutions, soda water, etc. In a solution, each element appears as a separate phase.

The ratio between the mass and the amount of a chemical compound's constituents is known as the compound's molar mass. A substance's molar mass is a bulk attribute rather than a molecular one.

π = cRT

c = n/V

n = w/m = 10*10^-3 /[m*30*10^-3]M

m = RT/383.14 = 18.22 kg/mol

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Answer:

180N

Explanation:

the gravity on Earth is six times the one on the moon

Answer:

I think 180N

Explanation:

the gravity of the surface of the earth is 6Times

more than the moon

Answer:

v₂ = 16 m/s

Explanation:

We can use the continuity equation, which is as follows:

[tex]A_1v_1 = A_2v_2\\[/tex]

where,

A₁ = Area of inlet = πd²/4

A₂ = Area of outlet = π(d/2)²/4 = πd²/16

v₁ = velocity at inlet = 4 m/s

v₂ = velocity at outlet = ?

Therefore,

[tex](\frac{\pi d^2}{4})(4\ m/s)=(\frac{\pi d^2}{16})v_2\\\\[/tex]

v₂ = 16 m/s

The correct answer is "Pulley C." In a system of three pulleys, where the girl is hanging from one end and the other end is fixed, the tension in the rope is equal throughout the system.

If a girl weighing 200 newtons hangs from three pulley systems, the reading on all three pulley systems would be 200 newtons. In an ideal pulley system, the tension in the rope is the same throughout, so the force applied to each pulley would be equal to the weight of the girl, which is 200 newtons in this case. The correct answer is "Pulley C." In a system of three pulleys, where the girl is hanging from one end and the other end is fixed, the tension in the rope is equal throughout the system.

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The example of functional fixedness is: B) Natasha doesn't think of using her CD case as an ice scraper to clear her windshield.

The example of functional fixedness is: B) Natasha doesn't think of using her CD case as an ice scraper to clear her windshield. Functional fixedness refers to a cognitive bias where an individual is unable to see alternative uses or functions for an object beyond its typical or intended purpose. In this case, Natasha is unable to think of using her CD case as an ice scraper, indicating functional fixedness.

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When ^227_89Ac undergoes alpha decay, the resulting nucleus is ^223_87Fr, with a decrease of 2 protons and 4 nucleons compared to Ac.

Alpha decay is a type of radioactive decay in which an atomic nucleus emits an alpha particle, which consists of two protons and two neutrons (equivalent to a helium nucleus). This process reduces the atomic number of the parent nucleus by 2 and its mass number by 4, resulting in the formation of a new nucleus. Alpha decay occurs in heavy, unstable nuclei to achieve greater stability by reducing their size and releasing excess energy.

When ^227_89Ac undergoes alpha decay, it emits an alpha particle, which consists of 2 protons and 2 neutrons. This means the resulting nucleus will have 2 fewer protons and 2 fewer neutrons compared to Ac.

Ac has an atomic number of 89, so after alpha decay, the resulting nucleus will have an atomic number of 89 - 2 = 87.

Ac has a mass number of 227, so the resulting nucleus will have a mass number of 227 - 4 = 223.

Therefore, the resulting nucleus is ^223_87Fr.

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The kinetic energy (KE) of the emitted photoelectrons in the Sodium surface will be: lower compared to the KE of the photoelectrons in the Potassium surface.

The work function of a material is the minimum amount of energy required to remove an electron from its surface. If the work function of Sodium is greater than that of Potassium, it means that Sodium requires more energy to remove electrons compared to Potassium.

When photons of the same wavelength are incident on both surfaces, the energy of the photons is given by E = hf, where h is Planck's constant and f is the frequency of the photons (related to the wavelength).

For the photoelectric effect to occur, the energy of the incident photons must exceed the work function of the material. Since Sodium has a higher work function than Potassium, it will require photons with higher energy to exceed its work function and emit photoelectrons.

Therefore, the photons incident on the Sodium surface, despite having the same wavelength as those incident on the Potassium surface, will have lower energy. As a result, the kinetic energy of the emitted photoelectrons in the Sodium surface will be lower compared to the kinetic energy of the photoelectrons in the Potassium surface.

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Answer:

D

Explanation:

Does not assosicate with Light

The angular magnification produced by the reflecting telescope when a 3.05 m focal length eyepiece is used is approximately -1.97.

To calculate the angular magnification produced by a telescope, we can use the formula

Angular Magnification = - (fobjective / feyepiece)

Where:

fobjective is the focal length of the objective mirror

feyepiece is the focal length of the eyepiece

In this case, the objective mirror has a radius of curvature of 12.0 m, so its focal length (fobjective) is half of the radius of curvature:

fobjective = 12.0 m / 2 = 6.0 m

The focal length of the eyepiece is given as 3.05 m (feyepiece).

Substituting the values into the formula:

Angular Magnification = -(6.0 m / 3.05 m)

Angular Magnification = -1.97

Since the angular magnification is negative, it indicates that the image produced by the telescope is inverted.

Therefore, the angular magnification produced by the reflecting telescope when a 3.05 m focal length eyepiece is used is approximately -1.97.

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