The more energy a wave carries the BLANK its amplitude.

José's lack of pain is likely due to the release of endorphins, which are natural pain relievers produced by the body during intense physical activity.

José's lack of pain after a long and bruising football game is most likely due to the release of endorphins. Endorphins are neurotransmitters produced by the central nervous system and the pituitary gland. Known as the body's natural pain relievers, they are released in response to stress or discomfort to inhibit the transmission of pain signals in the brain and produce a feeling of euphoria.

This phenomenon is often referred to as a 'runner's high' and can occur during and after intense physical activity.

Answer:

A). The nuclear regulatory commission.

Explanation:

I am taking the test rnn.

Flvs

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, [tex]\theta=20^0[/tex]

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

[tex]mg=2T\ sin\ \theta[/tex]

[tex]T=\dfrac{mg}{2\ sin\theta}[/tex]

[tex]T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}[/tex]

[tex]T=936.7\ N[/tex]

Hence, this is the required solution.

The tension in the line is [tex]\boxed{932.18{\text{ N}}}[/tex].

Further Explanation:

Free body diagram is the graphical representation or illustration of all the forces applied on the body with their directions. This helps us to visualize all the applied forces on the body with their directions and make us possible to solve complex problems based on kinematics.

Tension is the force exerted by the wire, cable or string like object when someone pulls it.

The force in the downward direction is due to gravity called as weight. Its magnitude depends upon the mass of the body and expressed as [tex]mg[/tex]. the term [tex]g[/tex] is called acceleration due to gravity and it has a constant value of [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex].

Given:

The mass of the student is [tex]65\text{ kg}[/tex].

The sag angle with respect to the horizontal is [tex]20^\circ[/tex].

Concept:

The student walking in the slack line exerts tension on the slack line ties with two trees. There is tension on the both side of the slack line. The horizontal component of tension both side of the slack line will be canceling each other effects and the vertical component of tension both side of the slack line is balanced by the weight of the student.

It can be expressed as:

[tex]2T\sin\theta=mg[/tex]

Here, [tex]T[/tex] is the tension, [tex]\theta[/tex] is the sag angle, [tex]m[/tex] is the mass of object and [tex]g[/tex] is the acceleration due to gravity.

Substitute [tex]20^\circ[/tex]  for [tex]T[/tex], [tex]65\text{ kg}[/tex] for [tex]m[/tex] and [tex]9.81{\text{ m/}}{{\text{s}}^2}[/tex] for [tex]g[/tex] in the above equation.

[tex]\begin{aligned}2T\sin 20^\circ&=\left( {65{\text{ kg}}}\right)\left( {9.81{\text{ m/}}{{\text{s}}^2}} \right) \hfill\\T&=\frac{{\left({65{\text{ kg}}} \right)\left( {9.81{\text{ m/}}{{\text{s}}^2}}\right)}}{{2\sin 20^\circ}}\hfill\\&=932.18{\text{ N}}\hfill\\ \end{aligned}[/tex]

Therefore, the tension in the line is [tex]\boxed{932.18{\text{ N}}}[/tex].

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Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

65 kg, student, walking, slackline, length, webbing, stretched, between, two, trees, sags, 20 deg, angle, relative, horizontal, tension, line, 932.18 N.

To supply each bulb with a voltage of 2.4 V, the value of resistor r should be 0.6 ohms.

To find the value of resistor r in order to supply each bulb with a voltage of 2.4 V, we can use the formula for voltage in a series circuit:

V = V1 + V2 + Vr

Since the voltage across each bulb should be 2.4 V, we have:

2.4 V = 2.4 V + 2.4 V + Vr

Simplifying the equation, we find that the value of resistor r should be 0.6 ohms.

Answer:

gravity

Explanation: Because when you are jumping

Answer:

Mass of the cart = 146 kg

Explanation:

A cart is pulled by a force of 250 N at an angle of 35° above the horizontal.

The cart accelerates at 1.4 m/s² horizontally.

Horizontal force = Fcosθ = 250 cos35° = 204.79N

We have F = ma

Substituting

        204.79 = m x 1.4  

              m = 146.28 kg = 146 kg

Mass of the cart = 146 kg

Answer:

146 kg

Explanation:

Correct on Edge 2021

The mass of a soccer ball traveling at a velocity of 50 m/s and has a kinetic energy of 500 J is 0.4kg.

The mass of a soccer ball can be calculated using the following formula:

K.E = ½mv²

Where;

According to this question, a soccer ball is traveling at a velocity of 50 m/s and has a kinetic energy of 500 J. The mass of the kinetic energy is as follows:

500 = ½ × m × 50²

500 = ½ × 2500m

500 = 1250m

m = 500/1250

m = 0.4kg

Therefore, the mass of a soccer ball traveling at a velocity of 50 m/s and has a kinetic energy of 500 J is 0.4kg.

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A. The component waves have different frequencies.

Answer:

1. Disagree; inertia is a property of matter to behave this way, not some kind of force.

Explanation:

Newton's first law of motion states that a body will remain in rest or in uniform motion while no external force is applied on it. What makes things to preserve their states of motion in inertia. It can be easy to understand remembering that the mass of the objects is the measure of their inertia. Since the mass is a property of matter so is the inertia.

Answer:

D

Explanation:

The correct answer is A. Scientists want to share measurements data that they can understand

Explanation:

In science, a standard system of measurement refers to a unified set or system of units that can be used internationally to measure data related to distance, temperature, energy, mass, etc. For example, the International System of Units establishes distance should be measured in meters rather than in inches, feet, etc. This guarantees all around the word scientist use the same units and can later share measurements despite language, culture or country because by using the same units scientists can understand data. Thus, a standard system of measurement is important because "scientists want to share measurements data that they can understand ".

10 m/s² to the left.

Given:

Question:

What the acceleration of the bear is?

This is the case for implementing Newton's second law, i.e., 'the resultant force is proportional to the acceleration'. If we also use SI units then 'the resultant force is equal to the product of the mass and the acceleration'.

[tex]\boxed{\boxed{ \ F_{net} = m \cdot a \ }}[/tex]

Let's calculate the net force.

Referring to the horizontal axis marked a positive to the right, then

[tex]\boxed{ \ F_{net} = 15 - 17 = - 2 \ N \ }[/tex]

A negative sign indicates that the bear is moving to the left.

We continue to calculate the bear's acceleration. Remember, the mass that is applied is the mass of the bear that is accelerating.

[tex]\boxed{ \ F_{net} = m \cdot a \rightarrow a = \frac{F_{net}}{m_{bear}} \ }[/tex]

[tex]\boxed{ \ a = \frac{2 \ kg.ms^{-2}}{0.2 \ ms^{-2}} \ }[/tex]

Thus, the magnitude and direction of the acceleration is [tex]\boxed{\boxed{ \ a = 10 \ m/s^2 \ to \ the \ left \ }}[/tex]

Keywords: two children, a 200 g stuffed bear, the 25 kg boy pulls to the right, a 15 n force, the 20 kg girl, left, a 17 n, what the acceleration, newton's second law, the magnitude and direction

The acceleration of the bear is [tex]\boxed{10{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}[/tex] and its direction of motion is towards left side.

Further explanation:

The net force acting on the bear is equal to the difference between force applied by the girl on the bear and force applied by boy on the bear.

Given:

The mass of the bear is [tex]200{\text{ g}}[/tex].

The force applied by the boy on the bear, towards right side, is [tex]15{\text{ N}}[/tex].

The force applied by the girl on the bear, towards left side, is [tex]17{\text{ N}}[/tex].

The mass of the boy is [tex]25{\text{ kg}}[/tex].

The mass of the girl is [tex]20{\text{ kg}}[/tex].

Formula and concept used:

From the Newton’s law of motion,

[tex]\sum F=ma[/tex]  

Here, [tex]\sum F[/tex] is the net force acting on the object, [tex]m[/tex] is the mass of object  and [tex]a[/tex] is the acceleration.

The weight of the boy and the girl will not affect the net force applied by them on the bear.

The net force applied on the bear is equal to the difference between force applied by the girl and force applied by the boy because the forces applied by the boy and girl are parallel to each other but opposite in the direction.

Therefore, from the vector law, the resultant of the parallel vectors acting  opposite in direction is equal to the difference between their magnitudes.  The direction of the resultant can be specified as the direction of the vector on larger magnitude.

The expression for the net force on the bear is:

[tex]\boxed{ma={F_g} - {F_b}}[/tex]                                                                            …… (1)

Here, [tex]{F_g}[/tex] is the force applied by the girl, [tex]{F_b}[/tex] is the force applied by the boy, [tex]m[/tex] is the mass of bear  and [tex]a[/tex] is the acceleration of the bear.

Thus, the acceleration of the bear is [tex]\boxed{10{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}}[/tex] and its direction of motion is towards left side.

Calculation:

Substitute the value of [tex]{F_g}[/tex], [tex]{F_b}[/tex] and [tex]m[/tex] in equation (1).

[tex]\begin{aligned}\frac{{200}}{{1000}}a&=17 - 15 \\0.2a&=2 \\a&=10{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}} \\ \end{aligned}[/tex]

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Answer detail:

Grade: College

Subject: Physics

Chapter: Dynamics of motion

Keywords:

Two children fighting, force on the bear, acceleration, newton's law, boy, girl, pulling bear towards left, 10m/s2, 10m/s^2, 10.0m/s2 , 10.0m/s^2, mass, net force, weight, magnitude and direction.

Explanation :  

The angular velocity of the planet in the elliptical orbit will vary. Its shows that the planet  travel slower when farther from the sun, then faster when closer to the sun.

1. Yes, Earth experiences the strongest gravitational force when it is closest to the Sun.

2. Yes,  Earth moves the greatest distance in thirty days when it is closest to the Sun.

3. Yes,  Earth travels with the slowest tangential speed when it is farthest from the Sun.

4. Yes,  Earth sweeps out the same area in the same time frame anywhere in its orbit

Kepler's second law : Kepler describe the motion of the planets around the sun. the line joining the planet and the sun sweeps out  the same area in the same time.

The characteristics of Earth’s orbit  that that agree and follow the pattern of Kepler’s second law are;

Earth experiences the strongest gravitational force when it is closest to the Sun.

Earth moves the greatest distance in thirty days when it is closest to the Sun.

Earth travels with the slowest tangential speed when it is farthest from the Sun.

Earth sweeps out the same area in the same time frame anywhere in its orbit

Therefore, Earth usually have strongest gravitational force as it moves close to the sun.

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Answer : The maximum acceleration will be 3.0 [tex] m/sec^{2} [/tex].

Explanation : We can calculate the force using the formula given below;

F = m . a ;

(F - Force; m - mass and a - acceleration);

given in the problem; mass (1) - 1200 kg; acceleration (1) - 4 [tex] m/sec^{2} [/tex]

So, on substituting we get,

F = [tex] m_{1} a_{1} [/tex] = 1200 X 4 = 4800 N; we get F = 4800 N

Now, when mass 400 kg is added then mass (2) will be 1600 kg and acceleration has to be found;

Here, the force remains the same; so it can be equated;

[tex] m_{1} a_{1} [/tex] = [tex] m_{2} a_{2} [/tex];

So, 1200 X 4 = (1200 + 400) x [tex] a_{2} [/tex]

Therefore, the answer will be [tex] a_{2} [/tex] = 3.0 [tex] m/sec^{2} [/tex]

ΔU = -Wint

Consdier the work of of interaction is W =m*g*h - equation -1

and the Potential energy U.

Final Potential energy Uf =0 , And the Initial Potential Energy Ui =m*g*h

Now we will write the equation for a Change in Potential energy ΔU,

ΔU = Uf - Ui

= 0-m*g*h

  ΔU = -m*g*h --Equation 2

Now compare the both equation

Wint = -ΔU

we can rewrite the above equation

ΔU = -W.

So our Answer is ΔU = -W. .

Answer:

3.71 × 10²¹ atoms

Explanation:

The molar mass of P₂O₅ is 283.89 g/mol. The moles of P₂O₅ corresponding to 0.250 g of P₂O₅ are:

0.250 g × (1 mol/283.89 g) = 8.81 × 10⁻⁴ mol

In 1 mole of P₂O₅ there are 6.02 × 10²³ molecules of P₂O₅ (Avogadro's number). In 8.81 × 10⁻⁴ moles of P₂O₅, the molecules of P₂O₅ are:

8.81 × 10⁻⁴ mol × (6.02 × 10²³ molecule/ 1 mol) = 5.30 × 10²⁰ molecule

In 1 molecule of P₂O₅, there are 7 atoms (2 atoms of P and 5 atoms of O). In 5.30 × 10²⁰ molecules of P₂O₅, the number of atoms is:

5.30 × 10²⁰ molecule × (7 atom/1 molecule) = 3.71 × 10²¹ atom

Final answer:

A person with a near point of 56.6cm is farsighted and needs convex lenses to correct their vision.

A person with a far point of 56.6cm is nearsighted and requires diverging lenses.

Explanation:

To correct for vision where a person has a near point of 56.6cm, they are likely farsighted. Farsightedness is corrected using convex lenses that converge the light before it hits the eye. This increases the eye's focusing power to bring nearby objects into clear vision.

For a person with a far point of 56.6cm, they are suffering from myopia, or nearsightedness. They require diverging lenses, typically negative diopter lenses, to spread out the light before it reaches the eye. This extends the far point of their vision, allowing them to see distant objects more clearly.

If the eyeglasses are held 1.50 cm from the eyes, we would need to consider the vertex distance in the lens formula to calculate the required lens power. The formula to determine the lens power needed, in diopters (D), is 1/f, where f is the focal length in meters. Since the eyeglasses are not in direct contact with the eye, the formula to find the effective focal length f' when considering vertex distance d is f' = f - d.

The normal force on a car traveling inside a vertical circular track differs at the top and bottom. At the top, the normal force is in the same direction as gravity, and at the bottom, it supports the car's weight against gravity plus provides the centripetal force. To find the normal force at the bottom, one must use the equations for circular motion taking into account the gravitational force.

To determine the normal force on the car at the bottom of the track (point A), we must consider that the car is in circular motion, and at the top and bottom of the track, the forces acting on the car are different due to its position relative to the center of the circular path.

At the top of the track, the normal force and the weight of the car both act downwards. Since the car is in circular motion and is not accelerating vertically, the net force must be equal to the centripetal force required to keep the car moving in a circle. This can be represented by the equation N + mg = mv2/r at point B, where N is the normal force, m is the mass of the car, g is the acceleration due to gravity, v is the speed of the car and r is the radius of the circular path.

At the bottom of the track, the normal force acts upwards while the weight of the car acts downwards. The normal force at the bottom must be greater than at the top because it must support the car's weight in addition to providing the centripetal force. So, the equation at point A is N - mg = mv2/r.

From the information provided, we know that the normal force at the top is 6.00 N, the mass of the car is 0.800 kg, and the gravitational acceleration is 9.8 m/s2. By writing out both equations for the normal force at the top and bottom, and substituting the known values, we can solve for the unknown normal force at the bottom of the track.

To determine how long it takes a person to breathe one mole of air at a rate of 77.0 ml/s, the ideal gas law and conversion to appropriate units would be necessary. However, the actual calculation was not provided, and thus the exact time in seconds cannot be given.

The question asks about the time required for a person at rest to breathe one mole of air given a breathing rate of 77.0 ml/s at conditions of 25 ℃ and 755 mmHg. To solve this, we need to employ the ideal gas law. The ideal gas law relates the volume (V), pressure (P), temperature (T), and the number of moles (n) of an ideal gas to the universal gas constant (R).

We use the conditions given by the student to calculate the volume of one mole of air under the specified conditions. The ideal gas law is PV = nRT where P is the pressure in atm, V is the volume in liters, n is the number of moles, R is the universal gas constant (0.0821 L·atm/K·mol), and T is the temperature in Kelvin.

First, convert the given pressure from mmHg to atm and the temperature from Celsius to Kelvin. Then solve for V when n is one mole.

However, without completing the calculation, we cannot provide the exact time needed as it depends on the precise volume one mole of air occupies at the given conditions. Once known, the time can be found by dividing this volume by the breathing rate (77.0 ml/s) and converting the result into seconds, which gives us the amount of time needed to breathe one mole of air.

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It would take approximately 318.2 seconds for a person at rest to breathe one mole of air, given the conditions of 25°C and 755 mmHg pressure. This calculation assumes ideal conditions and a constant breathing rate.

To determine how long it takes for a person at rest to breathe one mole of air, we need to calculate the volume of air that corresponds to one mole under the given conditions and then divide this by the breathing rate.

1. Calculate the volume of one mole of gas at STP (Standard Temperature and Pressure):

  - Standard temperature (T): 0°C = 273.15 K

  - Standard pressure (P): 1 atm = 760 mmHg = 101.325 kPa

  Using the ideal gas law [tex]\( PV = nRT \)[/tex]:

  \[ V_{\text{mole}} = \frac{nRT}{P} = \frac{(1 \text{ mol})(0.0821 \text{ L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1})(298.15 \text{ K})}{760 \text{ mmHg}} \approx 24.5 \text{ L} \][tex]\[ V_{\text{mole}} = \frac{nRT}{P} = \frac{(1 \text{ mol})(0.0821 \text{ L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1})(298.15 \text{ K})}{760 \text{ mmHg}} \approx 24.5 \text{ L} \][/tex]

2. Convert volume to milliliters (since the breathing rate is given in [tex]mL/s)[/tex]:

 [tex]\[ V_{\text{mole}} = 24.5 \text{ L} \times 1000 \text{ mL/L} = 24500 \text{ mL} \][/tex]

3. Calculate the time to breathe one mole of air:

[tex]\[ \text{Time} = \frac{V_{\text{mole}}}{\text{Breathing rate}} = \frac{24500 \text{ mL}}{77.0 \text{ mL/s}} \approx 318.2 \text{ seconds} \][/tex]

Therefore, it would take approximately 318.2 seconds for a person at rest to breathe one mole of air, given the conditions of 25°C and 755 mmHg pressure. This calculation assumes ideal conditions and a constant breathing rate.

Final answer:

The use of the mnemonic ROY G BIV to remember the colors of the rainbow in the order of wavelength illustrates the concept of color vision in physics.

Explanation:

The use of the mnemonic ROY G BIV to remember the colors of the rainbow in the order of wavelength illustrates the concept of color vision in physics. Visible light consists of different wavelengths, and our eyes perceive these wavelengths as different colors. The mnemonic helps us remember the order of colors in the rainbow from longest to shortest wavelength: red, orange, yellow, green, blue, indigo, and violet.

For example, red light has a longer wavelength than orange light, and violet light has a shorter wavelength than green light. The mnemonic ROY G BIV is a helpful tool for remembering this sequence of colors and their corresponding wavelengths.

Final answer:

The ratio of the horizontal distances of impact for the two water jets is 2.

Explanation:

To calculate the ratio of the horizontal distances of impact, we need to first find the time of flight for each water jet. The time of flight can be determined using the formula:

time = distance / velocity

For the first water jet with a height of 50 centimeters and a velocity of 1 meter/second, the time of flight is:

time 1 = 0.5 meters / 1 meter/second

= 0.5 seconds

For the second water jet with a height of 100 centimeters and a velocity of 0.5 meter/second, the time of flight is:

time 2 = 1 meter / 0.5 meter/second

= 2 seconds

Now, we can calculate the horizontal distances of impact using the formula:

distance = velocity * time

For the first water jet, the horizontal distance of impact is:

distance 1 = 1 meter/second * 0.5 seconds

= 0.5 meters

For the second water jet, the horizontal distance of impact is:

distance2 = 0.5 meter/second * 2 seconds = 1 meter

Therefore, the ratio of their horizontal distances of impact is:

ratio = distance 2 / distance 1

= 1 meter / 0.5 meters

= 2