The initial velocity of the skydiver is 0 ft/s. After 5 seconds, the velocity is approximately 34.5 ft/s. After 15 seconds, the velocity is approximately 60.2 ft/s.
Explanation:(a) To find the initial velocity of the skydiver, we need to evaluate v(t) at t = 0. Substitute t = 0 into the equation v(t) = A(1 − e^-kt) . Plugging in A = 64 and k = 0.19, we get v(0) = 64(1 - e^0) = 64(1 - 1) = 0 ft/s.
(b) To find the velocity after 5 seconds, substitute t = 5 into the equation. v(5) = 64(1 - e^(-0.19 * 5)) = 64(1 - e^(-0.95)) ≈ 34.5 ft/s .
To find the velocity after 15 seconds, substitute t = 15 into the equation. v(15) = 64(1 - e^(-0.19 * 15)) = 64(1 - e^(-2.85)) ≈ 60.2 ft/s .
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The initial velocity of the skydiver is 0 ft/s. After 5 seconds, the velocity is approximately 39.2 ft/s, and after 15 seconds, it is approximately 60.3 ft/s.
Let's solve the problem about the skydiver's velocity over time. The velocity function given is [tex]v(t) = A(1 - e^{-kt})[/tex] where A = 64 ft/s and k = 0.19 s⁻¹.
(a) Initial Velocity
The initial velocity of the skydiver occurs at t = 0. Plugging t = 0 into the velocity equation:
[tex]v(0) = 64(1 - e^{-0.19*0})[/tex]
Since e0 = 1:
v(0) = 64(1 − 1) = 0 ft/s
Therefore, the initial velocity of the skydiver is 0 ft/s.
(b) Velocity after 5 s and 15 s
Let's find the velocity after 5 seconds (t = 5):
[tex]v(5) = 64(1 - e^{-0.19*5})[/tex]
Calculating the exponent:
e-0.95 ≈ 0.387
So:
v(5) = 64(1 − 0.387) = 64(0.613) ≈ 39.2 ft/s
Now, let's find the velocity after 15 seconds (t = 15):
[tex]v(15) = 64(1 - e^{-0.19*15})[/tex]
Calculating the exponent:
e-2.85 ≈ 0.058
So:
v(15) = 64(1 − 0.058) = 64(0.942) ≈ 60.3 ft/s
Therefore, the velocities after 5 and 15 seconds are approximately 39.2 ft/s and 60.3 ft/s, respectively.
Where could convection currents form? Check all that apply.
in a sand dune
in a freshwater lake
in the atmosphere
in outer space
in Earth’s mantle
Answer:
In The Atmosphere
Explanation:
edg2021
The density of liquid oxygen at its boiling point is 1.14 \rm{kg/L} , and its heat of vaporization is 213 \rm{kJ/kg} .
How much energy in joules would be absorbed by 3.0 L of liquid oxygen as it vaporized?
To vaporize 3.0 L of liquid oxygen with a density of 1.14 kg/L and a heat of vaporization of 213 kJ/kg, 728460 joules of energy would be absorbed.
The student is asking about the amount of energy required to vaporize a certain volume of liquid oxygen. Given that the density of liquid oxygen at its boiling point is 1.14 kg/L, and its heat of vaporization is 213 kJ/kg, we can calculate the energy needed for vaporization using these two properties.
First, calculate the mass of 3.0 L of liquid oxygen:
Mass = Density times Volume = 1.14 kg/L times 3.0 L = 3.42 kg
Then, calculate the energy required for vaporization:
Energy = Mass times Heat of Vaporization = 3.42 kg times 213 kJ/kg = 728.46 kJ
Since 1 kJ = 1000 J, we can convert the energy to joules:
Energy in joules = 728.46 kJ times 1000 J/kJ = 728460 J
Therefore, 728460 joules of energy would be absorbed by 3.0 L of liquid oxygen as it vaporizes.
The Enlightenment period supported reason over religious beliefs.
Please select the best answer from the choices provided
T F
. An engineer wants to determine an efficient method for condensing large amounts of steam into liquid water. Which constant should she use?
A. - (triangle)H Vapor
B. - (triangle)H Fusion
The scientist should use the constant l (triangle)H Vapor, the Latent heat of vaporization.
What is Latent heat of vaporization, ◇Hvapor?The latent heat of vaporization is the amount of heat required to change a unit mass of a liquid to vapor at its boiling point and vice versa.
Since the scientist wants to determine an efficient method for condensing large amounts of steam into liquid water, he should use (triangle)H Vapor.
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Two identical small cars (car A & car B) have a head-on collision. Which scenario is true?
A. Car A exerts a greater force on car B than car B exerts on car A.
B. Car B exerts a greater force on car A than car A exerts on car B.
C. The force that car A exerts on car B and the force that car B exerts on car A are the same magnitude.
Answer:
The force that car A exerts on car B and the force that car B exerts on car A are the same magnitude.
Explanation:
There can be two types of collision i.e. elastic and inelastic collision. Elastic collision is also known as head-on collision. In this type of collision, the momentum and kinetic energy before and after the collision remains the same.
If two identical small cars (car A & car B) have a head-on collision, then the force that car A exerts on car B and the force that car B exerts on car A are the same magnitude. It is due to Newton's third law of motion which states that there is an equal and opposite reaction when one object applies a force on another object. So, the correct option is (c).
Jamal plugged his radio into the wall. The radio's plug had copper wires surrounded by rubber. The rubber protects Jamal from______?
me answer be ye Conduction
Dalton was one of the first scientists to experimentally prove that
a. the nucleus and the electron have different electrical charges.
b. most of the atom is made of "empty space".
c. the chemical and physical properties of an element correlate with its mass and size.
d. the atom is composed of smaller pieces.
Which of the following determine an object's velocity?
A.
speed and direction
B.
direction and acceleration
C.
speed and mass
D.
speed, direction, and acceleration
Speed and direction are the factors that affect an object's velocity.
What do your mean by velocity?The displacement that an object or particle experiences with respect to time is expressed vectorially as velocity. A vector quantity, that is. The change in an object's displacement with respect to time is known as velocity.
Displacement / Time = Velocity
The information is ,
Let V be the symbol for an object's velocity.
The measurement of V is now determined as
The concept of speed describes how quickly an object is traveling across a specific distance.
An object's speed just indicates how swiftly or slowly it is moving. It doesn't say which way the object is moving. Speed and direction are both referred to by the word velocity. A vector is an amount of velocity.
Now that the displacement is a vector quantity with direction, the equation for velocity is displacement / time.
As a result, an object's velocity is determined using its speed and direction.
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From Wien's law, at what wavelength does Jupiter's thermal emission peak?
µm In what part of the electromagnetic spectrum does this wavelength lie?
The density (mass/volume) of aluminum is 2.70 mc016-1.jpg 103 kilograms per cubic meter (kg/m3). What is the mass of an aluminum cylinder that has a volume of 1.50 m3?
The mass of the aluminum cylinder is 4,050 kg, obtained by multiplying the given density of aluminum with the given volume of the cylinder.
Explanation:The mass (m) of the aluminum cylinder can be calculated using the formula for density (ρ), which is Density = Mass/Volume. In this case, since the density of the aluminum (ϱ) is 2.70 * 103 kilograms per cubic meter (kg/m3) and the volume (V) of the cylinder is 1.50 m3, we can rearrange the formula to find mass such that Mass = Density * Volume. Substituting the given values, we get m = 2.70 * 103 kg/m3 * 1.50 m3 = 4,050 kg. Therefore, the mass of the aluminum cylinder is 4,050 kg.
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What causes the electric charges to flow from one end of the battery to the other?
a balance in electric potential
a balance in resistance
a difference in electric potential
a difference in resistance
Answer:
c. a difference in electric potential
Explanation:
Would you be doing any more work by going up the stairs twice as fast?
A 20,000 kg truck traveling at 25 m/s has a head-on inelastic collision with a 1500 kg car traveling at -30 m/s. calculate the initial momentum of the truck.
A 521-kg meteor is subject to a force of 2520 N. What is its acceleration?
Answer:
Acceleration, [tex]a=4.83\ m/s^2[/tex]
Explanation:
It is given that,
Mass of the meteor, m = 521 kg
Force acting on the meteor, F = 2520 N
Let a is the acceleration of the meteor. It can be calculated using the Newton's second law of motion. According to this law the force acting on an object is equal to the product of mass and acceleration with which it is moving. Mathematically, it is given by :
[tex]F=m\times a[/tex]
[tex]a=\dfrac{F}{m}[/tex]
[tex]a=\dfrac{2520\ N}{521\ kg}[/tex]
[tex]a=4.83\ m/s^2[/tex]
So, the acceleration of the meteor is [tex]4.83\ m/s^2[/tex]. Hence, this is the required solution.
In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.18 m. The mug slides off the counter and strikes the floor 1.20 m from the base of the counter. With what velocity did the mug leave the counter? What was the direction of the mug's velocity just before it hit the floor?
The distance traveled by an object divided by the time it takes to travel that distance is called
A. Average velocity
B. Average speed
C. Average acceleration
D. Negative acceleration
Energy and work are measured in the SI unit called
when a bullet is fired, does gunpowder only travels back towards the shooter
Answer:
false
Explanation:
Consider the observation the andromeda galaxy, a member of our local group, is moving toward us. why doesn’t this observation contradict the idea that the universe is expanding?
What is a measurement of the earth's history divided into time periods?
What are two ways that we use electromagnetic waves??
An object moves in uniform circular motion at 25 m/s and takes 1.0 second to go a quarter circle. What is the radius of the circle ? helpppppp me
Having some conceptual trouble with this problem: "A falling object travels one-fourth of its total distance in the last second of its fall. What height was it dropped from?" Would someone please help me set this one up?
Answer:
[tex]H = 273.4 m[/tex]
Explanation:
Let the falling object took "n" seconds to reach the ground and it travels H height
So we will have
[tex]H = \frac{1}{2}gn^2[/tex]
now we know that it covers one fourth of total height in last second
So we can say that it will cover 3H/4 distance in (n-1) seconds
so we will have
[tex]\frac{3H}{4} = \frac{1}{2}g(n-1)^2[/tex]
now from above two equations
[tex]\frac{4}{3} = (\frac{n}{n-1})^2[/tex]
[tex]1.155(n-1) = n[/tex]
[tex]n = 7.46 s[/tex]
Now we have
[tex]H = \frac{1}{2}(9.81)(7.46^2)[/tex]
[tex]H = 273.4 m[/tex]
If the work function for a certain metal is 1.8eV, what is the stopping potential for electrons ejected from the metal when light of wavelength 400nm shines on the metal? And what is the maximum speed of the ejected electrons? ...?
To find the stopping potential and the maximum speed of ejected electrons from a metal with a given work function when exposed to a specific wavelength of light, physicists use the photoelectric effect principles. Calculations involve the use of Planck's constant, the speed of light, and the electron charge to first determine the energy of incident photons and then the electrons' kinetic energy and velocity.
Explanation:To determine the stopping potential for electrons ejected from a metal when light of wavelength 400 nm shines on it, first, we use the photoelectric equation: E(Photon) = Work Function + Kinetic Energy(Max). The energy of a single photon (E) can be found with the equation E=hc/λ, where h is Planck's constant (6.626 x 10^-34 J·s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength in meters. For the stopping potential, the maximum kinetic energy of the ejected electrons is equal to the electrical energy qV, where q is the charge of an electron (1.602 x 10^-19 C) and V is the stopping potential.
Using these equations and the given work function, the maximum speed of electrons can be calculated using the formula for kinetic energy: KE = 0.5 * m * v^2, where m is the mass of the electron (9.109 x 10^-31 kg) and v is the velocity.
The exact calculations for the stopping potential and maximum speed are not provided here, since we are not certain about the correctness of these specific answers.
anyone know any of these please???
Forces contribute to the net force on a car rolling down a ramp.
a. Which force supports the car’s weight?
b. Which force accelerates the car down the ramp?
c. Which force acts against the motion of the car?
materiel in which the relative location of the atom is fixed are
Physics Help Please:
Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to trade. Astronaut 1 tosses the 0.130kg apple toward astronaut 2 with a speed of vi,1 = 1.07m/s . The 0.160kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.19m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 0.998m/s in the negative y direction.
What are the final speed and direction of the apple in this case?
I already found the speed (1.29m/s) but I cannot find the direction.
Picture is in the reply box.
Thank You
The problem involves applying conservation of momentum principles to a two-dimensional collision between an apple and an orange tossed by astronauts. The final direction of the apple after the collision can be determined using the velocity components and trigonometry, specifically the arctan function. To provide the direction, additional details like initial direction are needed.
Explanation:The student is seeking help with a physics problem involving the concepts of momentum conservation and two-dimensional collisions. In this problem, two astronauts (in a hypothetical scenario) are tossing an apple and an orange to each other when the fruits collide in space. The provided information lets us analyze the collision using the principles of momentum to find the final direction of the apple after the collision.
To solve for the direction of the apple after the collision, we must apply the law of conservation of momentum in two dimensions because the collision sends the objects off in different directions. Since there are no external forces, the total momentum in each direction (x and y) should remain constant. We use the before-collision and after-collision momenta to form equations and solve for the final velocity components of the apple, allowing us to calculate the final direction using trigonometry.
However, to provide the exact direction, a diagram or more information indicating the initial directions of the fruits would be required. Assuming the initial throw was along the x-axis and the final velocity of the orange is given in the y-axis, you can apply the arctan function to the velocity components of the apple to find the direction in terms of angle from the x-axis.
a man drops a ball downside from the roof of a tower of height 400 meters.At the same time another ball is thrown upside with a velocity 50 m/s from the surface of the tower,find when and at which height from the surface of the tower the two balls meet together.
The two balls meet at approximately 272.5 meters above the surface of the tower after around 5.1 seconds.
To determine when and at what height the two balls meet, we'll analyze their motions individually and then together.
Step 1: Motion of the dropped ball
For the ball dropped from the roof at a height of 400 meters, the equation of motion under gravity is:
h₁ = 400 - 0.5 * g * t²
where:
h₁ is the distance traveled by the dropped ball in meters.
g is the acceleration due to gravity (9.8 m/s²).
t is the time in seconds.
Putting in the values, h₁ = 400 - 4.9 t²
Step 2: Motion of the thrown ball
For the ball thrown upwards with an initial velocity of 50 m/s, the equation of motion is:
h₂ = 50t - 0.5 * g * t²
where:
h is the distance traveled by the thrown ball in meters.
Putting in the values, h₂ = 50t - 4.9 t²
Step 3: Equating the distances
The balls meet when their total distance is 400 meters, so:
h₁ + h₂ = 400
Substituting the equations for h₁ and h₂, 400 - 4.9 t² + 50t - 4.9 t² = 400
Simplifying, -9.8 t² + 50t = 0
Factoring out t, t(50 - 9.8t) = 0
Thus, t = 0 (initial point, not useful) or t = 50 / 9.8 ≈ 5.1 seconds.
Step 4: Calculating the height
Using t ≈ 5.1 seconds in any height equation:
h₁ = 400 - 4.9 * (5.1)²
h₁ = 400 - 127.5
h₁ ≈ 272.5 meters
Therefore, the two balls meet approximately 272.5 meters above the surface of the tower after about 5.1 seconds.
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