The balls are released from rest at a height of 5.0 m
at time t=0s
. Using these numbers and basic kinematics, you can determine the amount of time it takes for the balls to reach the ground.

Answers

Answer 1

Time taken by the ball to reach the ground is  1.0 second.

What is acceleration?

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

The initial height of the  balls = 5.0 meter

Initial time = 0 second

The initial speed of the ball = 01 m/s

Acceleration due to gravity = 10.0m/s²

Hence, Time taken by the ball to reach the ground is = √(2 × 5.0/10.0) second

= 1.0 second.

Therefor it take 1.0 second to ball

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Related Questions

The following circuit diagram is partially incomplete.
The current in Lamp 1 is 0.4 A. The potential difference across Lamp 1 is 8.0 V.
Calculate the resistance of Lamp 1.
Group of answer choices

A) 30 ohms
B) 20 ohms
C) 2 ohms
D) 3 ohms

Answers

Answer:

20Ω

Explanation:

we are here given that,

Potential difference= 8Vcurrent= 0.4 A Resistance= ?

From Ohm's law ,

[tex]\implies V = iR \\[/tex]

where,

V is potential differencei is currentR is resistance

on substituting the respective values, we have,

[tex]\implies 8V = 0.4A \times R\\[/tex]

[tex]\implies R =\dfrac{8V}{0.4A} \\[/tex]

[tex]\implies \underline{\underline{ R = 20\Omega}}\\[/tex]

and we are done!

A circuit contains an electrical power supply, a lamp, an ammeter, and a variable resistor. The resistance of the variable resistor is increased. What happens to the ammeter reading, and what happens to the brightness of the lamp?

A) The ammeter reading decreases, and the lamp's brightness increases.
B) The ammeter reading increases, and the lamp's brightness increases.
C) The ammeter reading decreases, and the lamp's brightness decreases.
D) The ammeter reading increases, and the lamp's brightness decreases.

Answers

b because the lamp light increases

A positive charge moves in the direction of an electric field. Which of the following statements are true?
Check all that apply.
A positive charge moves in the direction of an electric field. Which of the following statements are true?Check all that apply.
The potential energy associated with the charge decreases.
The electric field does not do any work on the charge.
The electric field does positive work on the charge.
The amount of work done on the charge cannot be determined without additional information.
The potential energy associated with the charge increases.
The electric field does negative work on the charge.

Answers

The first and third statements for the positive charge which moves in the electric field are correct.

When placed in an electric field, the positive charge typically flows in the direction of the electric field. The work done by the electric field in this situation is positive because both the electric field vector and the displacement vector point in the same direction and there is no angle between them.

The potential will also decrease in the direction of the electric field. This is because the work done in the order of the electric field is called the potential. electric potential is the amount of work required to move a unit charge from a reference point against an electric field to a particular point 

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Can someone answer this pls ?

Answers

It is discovered that the object, which weighs 60g and has a volume of 50cm³, has a density of 1.2 g/cm³.

Does density vary as a result of pressure?

Under pressure, a liquid becomes denser. In other words, as an excessive force is given to a liquid, volume decreases and density increases, and as molecules get closer to one another, interparticle spacing decreases.

What does mass vs. density mean?

A substance or object's mass is a measure of how much matter it contains. Volume (the amount of space an object or substance occupies) to mass (the amount of material present) ratio is known as density (its mass).

According to the given information:

Mass= 60g

Volume = 50cm³

Density = mass / volume

        => 60/50 =>1.2 g/cm³

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The motions described in each of the questions take place at an intersection on a two-lane road with a stop sign in each direction. For each motion, select the correct position versus time graph. For all of the motions, the stop sign is at the position x=0, and east is the positive x direction.

Answers

A) Motion 1: The car starts at rest at the stop sign, at x=0, and then accelerates in the east (positive x) direction.

What is accelerates?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity, meaning that it has both magnitude and direction. Acceleration is usually measured in meters per second squared (m/s2). It is a fundamental concept in physics, and is used to describe the motion of objects in a variety of contexts, from astronomy to everyday life. Acceleration can result from a force, from a change in direction, or from a change in speed.

The correct position vs time graph for this motion is a line with a positive slope, starting at the origin and going up to the right. This shows that the car is accelerating in the positive x direction and is increasing its position over time.

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A skier starts down a 10 degree incline at 2 m/s, reaching a speed of 15m/s at the bottom. What is the length of the incline?
How long does it take the skier to reach the bottom.

Answers

Answer:

-11.29 m

Explanation:

The length of the incline can be calculated using the following formula:

L = (Vf^2 - Vi^2) / (2 * a)

where:

L = length of incline

Vf = final speed (15 m/s)

Vi = initial speed (2 m/s)

a = acceleration due to gravity (9.8 m/s^2 on the surface of the Earth)

Since the skier is going downhill, the acceleration is negative.

Substituting the given values into the formula:

L = (15^2 - 2^2) / (2 * -9.8)

L = (225 - 4) / -19.6

L = 221 / -19.6

L = -11.29 m

So the length of the incline is approximately -11.29 meters. This negative value indicates that the skier's final position is lower than the starting position.

Transverse waves on a string have wave speed 12m/s, amplitude 0.05m and wavelength 0.4m. The waves travel in the positive x direction, and at t=0 the x = 0 end of the string has zero displacement and is moving upward.
(a) Find the frequency, period and wave number of these waves
(b)Write a wave function describing the wave
(c) Find the transverse displacement of a wave at x=0.25 at time t -0.15sec
(d) How much time must elapse from the instant in part "(c)" until the point at x = 0.25m has zero displacement​

Answers

(a) The wave speed (v) is given as 12m/s and the wavelength (λ) is 0.4m, so the frequency (f) can be calculated as:

v = fλ

f = v/λ = 12/0.4 = 30 Hz

The period (T) is the inverse of the frequency, so:

T = 1/f = 1/30 = 0.0333 s

The wave number (k) can be calculated as:

k = 2π/λ = 2π/0.4 = 15.708 rad/m

(b) The wave function can be written as:

y(x, t) = A sin(kx - ωt + φ)

where A is the amplitude, ω is the angular frequency, and φ is the phase constant. Since the x = 0 end of the string has zero displacement and is moving upward at t = 0, the phase constant is 0. At t = 0, the wave function can be written as:

y(x, 0) = A sin(kx)

The amplitude is given as 0.05m, so:

0.05 = A sin(k*0)

sin(0) = 0, so A = 0.05. Therefore, the wave function is:

y(x, t) = 0.05 sin(15.708x - 188.5t)

(c) To find the transverse displacement of the wave at x = 0.25m and t = -0.15s, we can substitute these values into the wave function:

y(0.25, -0.15) = 0.05 sin(15.7080.25 - 188.5(-0.15))

y(0.25, -0.15) = 0.05 sin(0.5906)

y(0.25, -0.15) ≈ 0.029 m

Therefore, the transverse displacement of the wave at x = 0.25m and t = -0.15s is approximately 0.029m.

(d) The point at x = 0.25m has zero displacement when the argument of the sine function is an integer multiple of π, or:

15.708(0.25) - 188.5t + φ = nπ

where n is an integer. We want to find the time at which this equation is satisfied. Rearranging, we get:

t = (15.708*0.25 + φ - nπ)/188.5

The phase constant is 0, so we can simplify the equation to:

t = (15.708*0.25 - nπ)/188.5

For n = 0, the time is:

t = (15.708*0.25 - 0)/188.5 ≈ 0.0208 s

Therefore, the time required for the point at x = 0.25m to have zero displacement is approximately 0.0208s.

Find the work W done if a constant force of 85 lb is used to pull a cart a distance of 220 ft. W=_____ft-lb

Answers

If a constant force of 85 lb is used to pull a cart a distance of 220 ft. Then the work done W is 18700 ft-lb.

Work is the amount of energy transferred when a force is applied over a distance. In this case, a constant force of 85 lb is used to pull a cart a distance of 220 ft. To find the work done, we can use the formula W = Fd, where F is the force and d is the distance. Plugging in the values, we get W = 85 x 220 = 18700 ft-lb. This is the amount of work done by the force to move the cart a distance of 220 ft.

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Two ice skaters stand together as illustrated in Figure (a) below. They "push off" and travel directly away from each other, the boy with a velocity of v = 0.540 m/s to the left. If the boy weighs 747 N and the girl weighs 497 N, what is the girl's velocity (in m/s) after they push off? (Consider the ice to be frictionless.)

Answers

the girl's velocity after they push off is 0.816 m/s.

What is velocity?

Velocity is described as the directional speed of an object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

The momentum of the boy after he pushes off is given by m_boy * v_boy = -747 N * 0.540 m/s = -405.08 N * m/s.

The momentum of the girl after she pushes off is denoted as   m_girl * v_girl, where m_girl is the mass of the girl and v_girl is her velocity.

The principle of conservation of momentum requires that the total momentum of the two-person system before and after the push off must be the same, hence we have:

m_boy * v_boy + m_girl * v_girl = 0

Substituting in the values for m_boy and v_boy, we find:

-405.08 N * m/s + m_girl * v_girl = 0

Solving for v_girl, we find:

v_girl = 405.08 N * m/s / m_girl = 405.08 N * m/s / 497 N = 0.816 m/s.

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Horsepower is defined as ______.
A. power applied over the distance of an English horserace track
B. the mechanical advantage gained by using a typical horse
C. the mass of a horse divided by the time it takes the horse to run 100 meters
D. a specific amount of power (550 ft-lb/s)

Answers

The definition of horsepower is . A. the amount of force used to cover the length of an English associated activities track B. the practical benefit of utilizing a standard horse

Who developed the horsepower?

James Watt, an engineer, is credited with creating the term horsepower. According to legend, Watt wanted to find a method to describe the power that one of those animals could produce while he was dealing with ponies in a coal mine. An engine is connected to a dynamometer in order to determine its horsepower.

What is the power of a horse?

The most prevalent unit of power is horsepower , which measures how quickly work is completed. According to the British Imperial Units, a horsepower is equivalent to 33,000 walking of work per minute, or the force required to elevate a mass of 3 million pounds one foot in a minute.

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when monkey mo is suspended at rest by holding a rope with one hand and the side of his cage with the other, all the force vectors that act on him

Answers

Monkey Mo is in a static equilibrium because of the balance of all these forces acting on him.

What is force?

The interaction between two things or the pressure applied to alter an object's motion is described by the physical quantity known as force. An object can change direction, accelerate, decelerate, or continue in motion as a result of force.

The forces at work on Monkey Mo when he is hung at rest while grasping a rope in one hand and the side of his cage in the other are as follows:

Gravitational force, which is vertically downward operating force of attraction between Monkey Mo and the Earth.

The tension force, which acts upward along the length of the rope, is the force the rope applies to Monkey Mo.

Normal force, which force balances the gravitational pull by operating perpendicular to the surface of the cage and being applied to Monkey Mo by the side of the cage.

Frictional force prevents Monkey Mo from falling out of the cage by acting between his hand and the side of the cage.

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a satellite of mass m moves in a circular orbit of radius r at a constant speed v. which of the following must be true? i. the net force on the satellite is equal to mv2 /r and is directed toward the center of the orbit. ii. the net work done on the satellite by gravity in one revolution is zero. iii. ithe angular momentum of the satellite is a constant.
(A) I only (B) III only (C) I and II only (D) II and III only (E) I, II, and III

Answers

The answer is (E) I, II, and III. When a satellite of mass m moves in a circular orbit of radius r at a constant speed v, all three statements are true.

i. The net force on the satellite is equal to mv^2 / r and is directed toward the center of the orbit. This force is known as the centripetal force, and it keeps the satellite moving in a circular path.

ii. The net work done on the satellite by gravity in one revolution is zero. Work is defined as the dot product of force and displacement, and the displacement of the satellite in a circular path is zero, so the net work done on it is also zero.

iii. The angular momentum of the satellite is a constant. Angular momentum is a measure of an object's rotational motion and is conserved if there are no external torques acting on the system. In this case, the satellite is in a stable orbit and there are no external torques acting on it, so its angular momentum is conserved and remains constant.

Therefore, all three statements (I, II, and III) must be true.

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A sphere and a cylinder of equal mass and radius are simultaneously released from rest on the same inclined plane and roll without sliding down the incline. Then: A. the sphere reaches the bottom first because it has the greater inertia B. the cylinder reaches the bottom first because it picks up more rotational energy C. the sphere reaches the bottom first because it picks up more rotational energy D. they reach the bottom together E. none of the above are true

Answers

The correct option for the problem is option D. they reach the bottom together. Their gravitational potential energy is converted to kinetic energy, total energy will be conserved for the system and we can safely say they will reach the bottom together without considering their shape.

Here sphere and cylinder both have equal mass and equal radius and it is given that both are simultaneously released also.

When two objects with equal mass and equal radius like these two, the time taken for them to roll down and reach the bottom will be their gravitational potential energy.

Here what happens is that their gravitational potential energy is converted to kinetic-energy and both of them rolling down the incline.

Total energy will be conserved for the system and we can safely say they will reach the bottom together without considering their shape.

Inertia and rotational energy are also there in this context for their motion but they do not take part any role for the time taken to slide down.

Instead, they affect the stability and trajectory of the objects as they roll down the incline.

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Which is NOT a function of the skeletal system?

Answers

Answer:produces red and white blood cells

Explanation: the bone marrow helps with the blood but cannot produce red and white blood cells

Answer:

"prevents movement of limbs and digits" is NOT a function of the skeletal system.

The condition for the photodynamic effect to occur is that the pigment must have
a. Phosphorescence
b. Fluorescence
c. The absorption spectrum is the same as the absorption spectrum of the substrate

Answers

The condition for the photodynamic effect to occur is that the pigment must have the absorption spectrum is the same as the absorption spectrum of the substrate.

What is  photodynamic treatment ?

In order to induce cell death  (phototoxicity), photodynamic treatment (PDT), a type of phototherapy, uses light, a photosensitizing agent, and molecular oxygen.

PDT is frequently used to treat acne. It is clinically used to treat a variety of medical problems, such as wet age-related macular degeneration, psoriasis, and atherosclerosis. It has also showed some promise in the treatment of viral diseases like herpes. Malignant malignancies of the head and neck, bladder, lung, and specific skin types are also treated by it.

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at a distance r1 from a point charge, the magnitude of the electric field created by the charge is 397 n/c. at a distance r2 from the charge, the field has a magnitude of 145 n/c. find the ratio r2/r1.

Answers

The distance ratio r2/r1 is approximately 0.605. 

How to calculate the ratio?

The point charge is given by

E = k*q/r^2

where E is the electric field, k is the Coulomb constant ([tex]k = 8.99 * 10^9 Nm^2/C^2[/tex]), q is the charge of the point charge, and r is the distance between the point charge and the location. The electric field is measured.

Using this equation, we can find the distance ratio r2/r1.

[tex]E1 = k*q/r1^2\\E2 = k*q/r2^2[/tex]

Dividing these equations gives:

[tex]E2/E1 = (kq/r2^2) / (kq/r1^2)\\E2/E1 = (r1/r2)^2[/tex]

Solving for r2/r1 gives:

r2/r1 = √(E2/E1)

Inserting the given value will result in:

r2/r1 = √(145n/C ÷ 397n/C)

r2/r1 = √(0.365)

r2/r1 = 0.605

Therefore, the distance ratio r2/r1 is approximately 0.605. 

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In your composition for a love story, how could you use tempo to reflect the mood when the ice cream van interrupted the wedding ceremony?
A. Slow, legato
B. Fast, upbeat
C. Ritardando

Answers

A. Slow, legato. for a love story, we use slow and legato tempo to reflect the mood when the ice cream van interrupted the wedding ceremony.

How does tempo appear in a score?

The beats per minute are the most precise way for a composer to specify the intended speed (BPM). This means that a specific note value is designated as the beat (for instance, a quarter note), and the marking specifies that a specific number of these beats must be performed each minute.

How do you determine a song's tempo?

The music's performance speed is thus determined by the tempo. So you can rapidly establish the Beats Per Minute, or BPM, by counting how many beats there are in a music played at a particular rate. If you're pushed for time, you can also double the amount of beats in 15 seconds of music by four.

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The 20-g centrifuge at NASA's Ames Research Center in Mountain View, California, is a horizontal, cylindrical tube of length 58.0 ft. Another 20-g centrifuge of length L = 59.0 ft is represented in the figure below. Assume an astronaut in training sits in a seat at one end, facing the axis of rotation 29.5 ft away. Determine the rotation rate, in revolutions per second, required to give the astronaut a centripetal acceleration of 20.0g. L/2. Part 1 of 4 - Conceptualize: Look carefully at the figure and imagine you are the astronaut sitting in the seat at the far right. As the centrifuge rotates, you are experiencing an acceleration toward the center of the device. The faster the centrifuge spins, the larger is that acceleration. The problem is asking for a rotation rate, measured in revolutions per second. Because the period of circular motion is the number of seconds per revolution, we see that the rotation rate will be the inverse of the period. rotation rate. 3) Based on the correct choice in question (2), substitute numerical values, including a unit conversion factor, to find the rotation rate for the centrifuge that will provide a centripetal acceleration of 20.0g for the astronaut. 1 ac 20.0( 9.8 m/s²), 1 1 X9 ft 2n 29.5 ft 1 m 0.74 rev/s %

Answers

For part 4 after solving the equation the the rotation rate is 0.74 rev/s.

What is rotation rate?

Rotation rate is the angular velocity of a body or object in a circular motion. It is typically measured in rotations per minute (RPM) and is the number of times a full rotation is completed in one minute.

Part 1 of 4 - Conceptualize: Look carefully at the figure and imagine you are the astronaut sitting in the seat at the far right. As the centrifuge rotates, you are experiencing an acceleration toward the center of the device.

The faster the centrifuge spins, the larger is that acceleration. The problem is asking for a rotation rate, measured in revolutions per second. Because the period of circular motion is the number of seconds per revolution, we see that the rotation rate will be the inverse of the period.

Part 2 of 4 - Calculate: To calculate the rotation rate, we need to find the period of the motion. The period of circular motion is given by:

T = 2π√(L/2a)

Where L is the length of the centrifuge and a is the centripetal acceleration.

Part 3 of 4 - Calculate: In this problem, L = 59.0 ft, and a = 20.0g. Since gravitational acceleration is equal to 9.8 m/s2, we need to convert the acceleration from g to m/s2. We can do this by multiplying the acceleration by 9.8. Thus, a = 20.0g × 9.8 m/s2 = 196.0 m/s2. Substituting the numerical values into the equation, we get:

T = 2π√(59.0 ft/2 × 196.0 m/s2) = 2π√(29.5 ft × 9.8 m/s2) = 2π√(291.1 m2/s2) = 2π√(291.1) s = 5.15 s

Part 4 of 4 - Calculate: Since the period of circular motion is the number of seconds per revolution, the rotation rate is equal to the inverse of the period. Thus, the rotation rate is:

Rotation rate = 1/T = 1/5.15 s = 0.74 rev/s

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When connected to a 12V battery, the current in a car headlight is 4.0 A.
When the lamp is lit, what is its resistance?

A) 3.0 ohms
B) 16 ohms
C) 48 ohms
D) 0.33 ohms

Answers

Answer:

A) 3.0 ohms

Explanation:

R=V/I=12V/4.0A=3.0 ohms

Two (unnormalized) excited state wavefunctions of the H atom are (i) psi= (2-r/a)*e^(-r/2a) and (ii) psi=r*sin(theta)*cos(phi)*e^(-r/2a).(a). Normalize both functions to one and confirm that these two functions are mutually orthogonal.(b) Evaluate the expectation values of r and r^2 for the atom.

Answers

(a) Normalization factor for wavefunction 1 is, [tex]A = \sqrt{\dfrac{3^3}{{(16\pi a^5)}}[/tex], for wavefunction 2 is, [tex]B = \sqrt{\dfrac{15}{16\pi a^5}}[/tex]. The two wavefunctions are mutually orthogonal. (b) The expectation value of r for [tex]\psi_1[/tex] is 32.

(a) To normalize the wavefunctions, we need to find the normalization constants A and B such that

[tex]\int \|{\psi_1}\|^2 dV = 1[/tex] and [tex]\int \|{\psi_2}\|^2 dV = 1[/tex]

where [tex]\psi_1[/tex] and [tex]\psi_2[/tex] are the two wavefunctions given.

(i) [tex]\psi_1 = (2 - \dfrac{r}{a}) \times e^{\dfrac{-r}{2a}}[/tex]

The normalization condition for psi1 is:

[tex]\int \|\psi_1\|^2 dV = \int \{2 - \dfrac{r}{a} \times e^{\dfrac{-r}{a}}\}^2 r^2 sin\theta dr d\theta d\phi\\\\ = 1[/tex]

Evaluating the integral using spherical coordinates, we get:

[tex]\int_0^\infty \int_0^\pi \int_0^{2\pi} [(2 - r/a)^2 e^{-r/a} r^2 sin\theta]\ dr\ d\theta\ d\phi = 1\\\\\dfrac{(16\pi a^5)}{(3^3)} \times \|A\|^2 = 1\\A = \dfrac{3^3}{{(16\pi a^5)}^{0.5}}[/tex]

(ii) [tex]\psi_2 = r sin\theta\ cos\phi\ e^{\dfrac{-r}{2a}}[/tex]

Evaluating the integral using spherical coordinates, we get:

[tex]\int_0^\infty \int_0^\pi \int_0^{2\pi} r^4 sin^3\theta cos^2\phi e^{-r/a} dr\ d\theta\ d\phi = 1\\\\\dfrac{16\pi a^5}{15} \times \|B\|^2 = 1\\B = \sqrt{\dfrac{15}{16\pi a^5}}[/tex]

To confirm that the two wavefunctions are mutually orthogonal, we need to evaluate the integral [tex]\int \psi_1\times \psi_2 dV[/tex] and show that it is equal to zero. Using spherical coordinates, we get:

[tex]\int_0^\infty \int_0^\pi \int_0^{2\pi} [(2 - r/a) \times e^{-r/a} \times r sin\theta cos\phi \times e^{-r/2a}]\ r^2\ sin\theta\ dr\ d\theta\ d\phi\\\\= 8\pi a^5 \times [\int_0^\infty e^{-r/a} r^2 (2 - r/a) e^{-r/2a}\ dr] \times [\int_0^\pi sin\theta\ cos\theta\ d\theta] \times [\int_0^{2\pi} cos\phi\ d\phi]\\= 0[/tex]

Therefore, the two wavefunctions are mutually orthogonal.

(b) To evaluate the expectation values of r and r^2, we need to calculate the integrals [tex]\int psi_1 \times r\ \psi_1 dV[/tex] and [tex]\int \psi_1 \times r^2\ \psi_1 dV[/tex]. Using spherical coordinates, we get:

For psi1:

[tex]\int psi_1 \times r\ \psi_1 dV = 4\pi \int_0^\infty (2 - r/a)^2 e^{-r/a} r^4 dr\\\\ = \dfrac{32\pi a^5}{15}[/tex]

[tex]\int \psi_1 \times r^2\ \psi_1 dV = 4\pi \int (2 - r/a)^2 e^{-r/a} r^5 dr\\\\ = \dfrac{128\pi a^6}{45}[/tex]

Therefore, the expectation value of r for [tex]\psi_1[/tex] is 32.

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Using Thomson's ___________ and Millikan's _____, the _____ of an ________ was determined.

Answers

Using Thomson's cathode ray tube and Millikan's oil drop experiment, the charge of an electron was determined.

Thomson's cathode ray tube and Millikan's oil drop experiment were two important experiments in the history of physics. Thomson used his cathode ray tube to observe the behavior of electrons, while Millikan used his oil drop experiment to measure the charge of an electron. By combining the results of these two experiments, scientists were able to determine the charge of an electron, which is a fundamental unit of electrical charge in the study of electricity and electromagnetism.

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a tow rope pulls a skier directly behind a boat with a force of 25 N for a total of 20 meters. How much work has the rope done on the skier?

Answers

The amount of work done by the applied force on the skier is 500 J.

What is the amount of work done by the applied force?

The amount of work done by the applied force on the skier is calculated by applying the following equation as shown below.

W = Fd

where;

F is the applied forced is the displacement of the skier

The amount of work done by the applied force on the skier is calculated as;

W = (25 N ) x ( 20 m )

W = 500 J

Thus, the work done by a force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.

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An automobile tire has a volume of 0.0185 m³. At a temperature of 289 K the absolute pressure in the tire is 217 kPa. How many moles of air must be pumped into the tire to increase its pressure to 272 kPa, given that the temperature and volume of the tire remain constant?​

Answers

0.0018 moles of air must be pumped into the tire to increase its pressure to 272 kPa.

How to find the number of moles of air

We can use the ideal gas law to find the number of moles of air in the tire.

The ideal gas law states that PV = nRT,

where

P is pressure,

V is volume,

n is the number of moles of gas,

R is the ideal gas constant, and

T is temperature.

The initial number of moles of air in the tire:

n1 = (P1 * V) / (R * T)

= (217 kPa * 0.0185 m³) / (8.31 J/mol * K * 289 K)

= 0.00625 moles

Next, we can find the final number of moles of air in the tire, assuming the temperature and volume remain constant:

n2 = (P2 * V) / (R * T)

= (272 kPa * 0.0185 m³) / (8.31 J/mol * K * 289 K)

= 0.00805 moles

Finally, we can find the difference between the final and initial number of moles, which represents the number of moles of air that must be pumped into the tire:

n2 - n1 = 0.00805 moles - 0.00625 moles = 0.0018 moles.

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If the net force on a baseball with a mass of 0.145 kg is 18.9 N, then was is the resulting acceleration?

Answers

Answer:

130.34 m/s²

Explanation:

see attachment

hope this helps!

Calculate the frequency in hertz of electromagnetic radiation that has a wavelength of 536.0 nm. (c = 3.00 X 10⁸ m/s)

Answers

The frequency of electromagnetic radiation with a wavelength of 536.0 nm is 5.6 x 10¹¹ Hz.

What is electromagnetic radiation?

Electromagnetic radiation is a form of energy that consists of waves of electric and magnetic fields, travelling through the air or other substances at the speed of light. These waves are created when an electric charge is accelerated and can be generated from a variety of sources, such as a light bulb, the sun, a microwave oven, or a radio transmitter.

The frequency (f) of electromagnetic radiation is calculated using the equation:
f = c/λ
where c is the speed of light (3.00 x 10⁸ m/s) and λ is the wavelength of the radiation (536.0 nm).
Therefore, we can calculate the frequency (f) as follows:
f = (3.00 x 10⁸ m/s) / (536.0 nm)
First, we need to convert the wavelength from nanometers (nm) to meters (m). We can do this by dividing 536.0 nm by 1,000,000 to get 0.000536 m.
Therefore, we can calculate the frequency (f) as follows:
f = (3.00 x 10⁸ m/s) / (0.000536 m)
f = 5.6 x 10¹¹ Hz
The frequency of electromagnetic radiation with a wavelength of 536.0 nm is 5.6 x 10¹¹ Hz.

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0.50-kg cord is stretched between two supports, 7.3m apart. When one support is struck by a hammer, a transverse wave travels down the cord and reaches the other support in 0.72s . What is the tension in the cord?

Answers

The tension in the cord will be 6.99 N. To solve this problem we will refer to the laws of Mersenne. Mersenne's laws are regulations relating to the frequency of oscillation of a stretched string or monochord, helpful in musical tuning and musical instrument building.

This law suggests that the velocity in a string is directly proportional to the root of the applied tension and inversely proportional to the root of the linear density, that is,

v = √T/μ

Here,

v = Velocity

μ = Linear density ( mass per unit length)

T = Tension

Rearranging to find the Period we have that

T = v²μ

T = v²(m/L)

As we know that speed is equal to displacement in a unit of time, we will have to

T = (L/t)² (m/L)

T = (7.3/0.72)² (0.50/7.3)

T = 6.99 N

Therefore, the tension in the cord will be 6.99 N.

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Felipe is riding his skateboard toward Lilah at 15 km/h. He throws a football to Lilah. The football is thrown at 8 km/h.

Answers

The velocity of the football that is thrown by Felipe to Lilah will be 23 km per hour.

What is the relative velocity?

The movement of an object in relation to another observer is known as its relative velocity. It is the pace at which one object's relative location changes in relation to another object over time.

Felipe is riding his skateboard toward Lilah at 15 km/h. He throws a football to Lilah. The football is thrown at 8 km/h.

Then the velocity of the ball is given as,

v = 15 + 8

v = 23 km/h

The velocity of the ball will be 23 km / h.

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A star of mass 5 × 10^30 kg is located at ‹ 8 × 10^12, 5 × 10^12, 0 › m. A planet of mass 6 × 10^24 kg is located at ‹ 5 × 10^12, 8 × 10^12, 0 › m and is moving with a velocity of ‹ 0.5 × 10^4, 1.5 × 10^4, 0 › m/s. During a time interval of 1×10^6 seconds, what is the change in the planet's velocity (Vf - Vi)?

Answers

The change in the velocity of the planet is 18.55 m/s.

What is the change in the planets velocity?

The gravitational force between the star and the planet, is calculated by applying Newton's third law of motion.

F = Gm₁m₂ /r²

where;

r is the distance between the star and the planet\m₁ is the mass of the starm₂ is the mass of the planet

The distance between the start and the planet is calculated as;

r = √ ( 8 x 10¹² - 5 x 10¹² )² + ( 5 x 10¹² - 8 x 10¹² )²

r = 4.24 x 10¹² m

F = ( 6.67 x 10⁻¹¹ x 5 x 10³⁰ x 6 x 10²⁴ ) / ( 4.24 x 10¹² )²

F = 1.11 x 10²⁰ N

The acceleration of the planet is calculated as;

a = F / m₂

a = ( 1.11 x 10²⁰ N ) / ( 6 x 10²⁴ kg )

a = 1.85 x 10⁻⁵ m/s²

The change in the velocity of the planet;

Δv = at

Δv = ( 1.85 x 10⁻⁵ x 1 x 10⁶ s )

Δv = 18.55 m/s

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A mass on a string of unknown length oscillates as a pendulum with a period of 3.3 s. What is the period if a. The mass is doubled?b. The string length is doubled?c. The string length is halved?d. The amplitude is doubled?

Answers

a. The mass is doubled = 4 s

b. The string length is doubled= 5.66 s

c. The string length is halved=2.83 s

d.The amplitude is doubled= 4 s

How to calculate period of pendulum

(a) The period of the pendulum is independent of the mass, therefore, the period when the mass is doubled is

T= To =4.00s

(b) Let I be the new length of the pendulum and Lo be the original length of the pendulum. The period of the pendulum if the string length is doubled is found by substituting 2Lo for L in Equation (*):

2π=√2Lo/g

√2=(2π(√Lo/g)

where the term in parenthesis is the original period of the pendulum To:

T=√√2To

T=√2.(4seconds)

T=5.66 s

c) The period of the pendulum if the string length is halved is found by substituting Lo/2 for L in Equation (*):

T=2π(√Lo/2 :g)

T= 1/√2 (2π√(Lo/g)

where the term in parenthesis is the original period of the pendulum To:

T =To /√2

T=4.00s/√2

T=2.83 s

d.) The period of the pendulum is independent of the amplitude, therefore, the period when the amplitude is halved is

T= To 4.00s

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earth moves about 30 km/s relative to the sun. when you jump upward in front of a wall, the wall doesn't slam into you at 30 km/s because the wall

Answers

The Earth's motion is unrelated to your motion, so the wall doesn't slam into you at 30 km/s.

What is Motion?

Motion is the change in an object's location with regard to its surroundings over a specific amount of time.

While the earth moves relative to the sun at a speed of around 30 km/s, this velocity has no bearing on how things move on the surface of the planet. Because the motion of things on the surface of the earth is governed by the gravity of the earth rather than its velocity relative to the sun, when you jump upward in front of a wall, the wall does not slam into you at a speed of 30 km/s.

The uniform gravitational field produced by the earth's gravity causes things on its surface to accelerate steadily in the direction of the planet's center. Your muscles generate an upward push that causes you to jump upward, leaving the ground.

Similar gravitational effects are also felt by the wall, which maintains its equilibrium with respect to the surface of the earth. Since no outside force is acting on the wall to cause it to shift position when you jump upward, the wall does not move towards you.

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