Answer:
C) upward
Explanation:
The problem can be solved by using the right-hand rule.
First of all, we notice at the location of the negatively charged particle (above the wire), the magnetic field produced by the wire points out of the page (because the current is to the right, so by using the right hand, putting the thumb to the right (as the current) and wrapping the other fingers around it, we see that the direction of the field above the wire is out of the page).
Now we can apply the right hand rule to the charged particle:
- index finger: velocity of the particle, to the right
- middle finger: direction of the magnetic field, out of the page
- thumb: direction of the force, downward --> however, the charge is negative, so we must reverse the direction --> upward
Therefore, the direction of the magnetic force is upward.
The magnetic force exerted on a negatively charged particle, moving in the same direction as the current, would be downward according to the left-hand rule in Physics.
Explanation:In this scenario, we can use the left-hand rule, as it's relevant to the movement and force applied on negatively charged particles in a magnetic field. To apply the left-hand rule, point your thumb in the direction of the particle's velocity (right), and your fingers in the direction of the current (also right). This should make your palm face downward. Hence, a negatively charged particle moving right, with the current also flowing right, would experience a magnetic force directed downward. So, the correct answer is B) Downward.
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As the ambulance got closer, Marge noticed that the pitch of the siren got higher. This happened because
A) the sound waves were pushed closer together.
B) the human ear increases pitch over time.
C) the ambulance was further away.
D) the volume increased.
A) the sound waves were pushed closer together
Answer:
The correct choice is
A) the sound waves were pushed closer together.
Explanation:
As the ambulance gets closer, the distance between the ambulance and marge decreases and so does the wavelength of the siren observed by marge. As the wavelength decrease and speed remains the same for the siren, the frequency of the siren increases because wavelength, frequency and speed are related as
Frequency = Speed/wavelength
Clearly, the frequency is inversely related to wavelength.
Electromagnetic force is present when electromagnetic fields
a- become charged as they lose electrons.
b- become stable as they come in contact with other forces.
c- interact with charged particles.
d- gain energy from other forces.
b is the correct answer
Answer:
c- interact with charged particles.
Explanation:
When a charged particle is placed in the field of the other charged particle then it will experience electromagnetic force on it.
As we know that force on a charged particle due to some other electric field is given by
F = qE
here we know
q = charge
E = electric field intensity
Similarly when a charged particle is placed in external magnetic field then the force on that moving charge is given by
[tex]F = q(v\times B)[/tex]
so here if charge is moving in the magnetic field due to some other system then the force on it is given by above equation.
So here we can say that electromagnetic force is present when electromagnetic fields
c- interact with charged particles.
How does an impulse propagate down the axon
When action potentials reach the end of the axon, they stimulate opening of Ca2+ channels, causing a release of neurotransmitters to the post-synaptic cell. How does and impulse propagate down the axon? The stimulus causes a start of the action potential and it moves down the axon without the ions moving down
A sound wave traveling in air has a frequency f and wavelength λ. A second sound wave traveling in air has wavelength λ/2. What is the frequency of the second sound wave?
Answer:
2f
Explanation:
The relationship between the speed of a sound wave (v), its frequency (f) and its wavelength is
[tex]v=\lambda f[/tex]
For the first wave speed, we have
[tex]v=\lambda f[/tex] (1)
Since the speed of a sound wave depends only on the medium, we can say that the second wave travels at same speed v (because it is still travelling in air, as the first wave). The wavelength of the second wave is [tex]\frac{\lambda}{2}[/tex], so if we call its frequency f', the new equation is
[tex]v=\frac{\lambda}{2}f'[/tex] (2)
And if we equate (1) and (2), we find
[tex]\lambda f = \frac{\lambda}{2}f'\\f'=2f[/tex]
The frequency of the second sound wave is 2f
[tex]\texttt{ }[/tex]
Further explanationLet's recall the speed of wave formula as follows:
[tex]\large {\boxed {v = \lambda f}}[/tex]
f = frequency of wave ( Hz )
v = speed of wave ( m/s )
λ = wavelength ( m )
Let's tackle the problem!
[tex]\texttt{ }[/tex]
Given:
frequency of first wave = f
wavelength of first wave = λ
wavelegnth of second wave = λ/2
Asked:
frequency of second wave = f' = ?
Solution:
Because the speed of wave depends on the medium, then:
[tex]\texttt{Speed of First Wave = Speed of Second Wave}[/tex]
[tex]v_1 = v_2[/tex]
[tex]\lambda_1 f_1 = \lambda_2 f_2[/tex]
[tex]\lambda f = \frac{1}{2}\lambda f'[/tex]
[tex]f = \frac{1}{2} f'[/tex]
[tex]f' = 2f[/tex]
[tex]\texttt{ }[/tex]
Conclusion:The frequency of the second sound wave is 2f
[tex]\texttt{ }[/tex]
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Answer detailsGrade: College
Subject: Physics
Chapter: Sound Waves
[tex]\texttt{ }[/tex]
Keywords: Sound, Wave , Wavelength , Doppler , Effect , Policeman , Stationary , Frequency , Speed , Beats, Medium, Space
What type of device is a car engine?
A.A heat pump that uses heat to do work
B.A heat engine that uses work to move heat
C.A heat pump that uses work to move heat
D.A heat engine that uses heat to do work
A heat engine that uses work to move heat
hope this helps :)
Answer:
D. A heat engine that uses heat to do work
Explanation:
Combustion = Heat
Combustion is used to create motion (work)
An astronaut has landed on planet n-40 and conducts an experiment to determine the acceleration due to gravity on that planet. She uses a simple pendulum that is 0.640 m long and measures that 10 complete oscillations 26.0 s. What is the acceleration of gravity on planet n-40?
Answer:
3.73 m/s^2
Explanation:
The period of a simple pendulum is given by
[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex] (1)
where
L is the length of the pendulum
g is the gravitational acceleration on the planet
The pendulum in the problem has length
L = 0.640 m
and makes 10 oscillations in 26.0 s; it means that its frequency is
[tex]f=\frac{10}{26 s}=0.385 Hz[/tex]
And so its period is
[tex]T=\frac{1}{f}=\frac{1}{0.385 Hz}=2.6 s[/tex]
So now we can solve equation (1) using L=0.640 m and T=2.6 s, so we can find the value of g on the planet:
[tex]g=(\frac{2\pi}{T})^2L=(\frac{2\pi}{2.6 s})^2 (0.640 m)=3.73 m/s^2[/tex]
What kind of wave is created if a tree falls in a forest?
sound waves even if no one is around to hear it.
A falling tree in a forest creates a sound wave. This is caused by the energy of the tree disturbing the atoms in the air, causing them to vibrate and create a wave. Sound can only be produced where there is matter like air or water.
Explanation:When a tree falls in a forest, it creates a disturbance in the air around it. This disturbance of air particles initiates a sound wave. Sound waves are a type of wave that our ears have evolved to detect.
As the tree makes impact, the energy transfers to the atoms in the air, causing them to vibrate. These vibrations then spread from particle to particle, creating a wave that travels through the air. This is the definition of sound in physics, regardless of whether there is anyone present to hear it or not.
However, it's important to note that these sound waves would only be produced in an environment that has matter (like air or water) for the wave to travel through. In space, where there is a vacuum, no sound would be produced if a tree fell because there would be no matter to disturb.
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When light goes from one material into another material having a HIGHER index of refractionA) its speed decreases but its wavelength and frequency both increaseB) its speed, wavelength, and frequency all decreaseC) its speed increases, its wavelength decreases, and its frequency stays the sameD) its speed decreases but its frequency and wavelength stay the sameE) its speed and wavelength decrease, but its frequency stays the same
Answer:
E) its speed and wavelength decrease, but its frequency stays the same
Explanation:
First of all, the frequency of a light wave does not depend on the medium, while wavelength and speed do. Therefore, the frequency remains costant.
In particular, the speed of light in a medium is given by:
[tex]v=\frac{c}{n}[/tex]
where c is the speed of light in a vacuum and n is the index of refraction. From the formula, we see that v and n are inversely proportional: so, when the light moves into a material with higher index of refraction, its speed decreases.
Moreover, speed is related to wavelength by
[tex]v=\lambda f[/tex]
where [tex]\lambda[/tex] is the wavelength and f is the frequency. Since the two quantities are directly proportional, this means that since the speed decreases, the wavelength decreases as well.
So, the correct choice is
E) its speed and wavelength decrease, but its frequency stays the same
When light goes from one material into another material having a HIGHER index of refraction , E) its speed and wavelength decrease, but its frequency stays the same.
[tex]\texttt{ }[/tex]
Further explanationLet's recall Snell's Law about Refraction as follows:
[tex]\boxed{n_1 \sin \theta_1 = n_2 \sin \theta_2}[/tex]
where:
n₁ = incident index
θ₁ = incident angle
n₂ = refracted index
θ₂ = refracted angle
[tex]\texttt{ }[/tex]
In this problem, we will use this following formula:
[tex]\boxed{ \frac{n_1}{n_2} = \frac{v_2}{v_1} }[/tex]
where:
n₁ = incident index
v₁ = incident speed
n₂ = refracted index
v₂ = refracted speed
[tex]\texttt{ }[/tex]
From above formula we could conclude that the speed of light is inversely proportional to the index of refraction. Therefore, when light goes from one material into another material having a HIGHER index of refraction , its speed will decrease.
[tex]\texttt{ }[/tex]
As we know that :
[tex]\boxed{v = \lambda f}[/tex]
Speed of light is directly proportional to the wavelength of light.
If the speed of light decreases , then wavelength of light will also decrease.
[tex]\texttt{ }[/tex]
Conclusion:When light goes from one material into another material having a HIGHER index of refraction , E) its speed and wavelength decrease, but its frequency stays the same
[tex]\texttt{ }[/tex]
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Answer detailsGrade: High School
Subject: Physics
Chapter: Light
A vector is 0.888 m long and points in a 205 degree direction.
Find the x-component of the vector.
Answer:
-0.805 m
Explanation:
The x-component of a vector is given by:
[tex]v_x = v cos \theta[/tex]
where
v is the magnitude of the vector
[tex]\theta[/tex] is the angle of the vector with respect to the positive x-direction
In this problem we have
v = 0.888 m
[tex]\theta=205^{\circ}[/tex]
so we have
[tex]v_x = (0.888 m)(cos 205^{\circ})=-0.805 m[/tex]
For a batter to increase momentum at the plate, he/she can? Choose a lighter bat Choose a heavier bat Choose an aluminum bat Choose a wooden bat
Choose a heavier bat
Explanation;Momentum is a vector which is often defined as the product of an object’s mass with its velocity.Two factors affecting momentum is the mass and velocity of the object. An object that has a low velocity and a small mass produce minimal momentum because it would take a small force and/or time to stop it.Therefore, choosing a heavier bat would increase the momentum of the plate.Final answer:
To boost momentum, a batter is suggested to choose a heavier bat due to its larger moment of inertia and the greater force it can impart on the ball, according to the principles of moment of inertia, angular momentum, and Newton's third law of motion.
Explanation:
To increase momentum at the plate, a batter can choose a heavier bat. This is because a more massive bat has more inertia and therefore, when swung, will impart a larger force on the ball for the same amount of time compared to a lighter bat. This is evident from the concepts of moment of inertia and angular momentum. When a bat is swung, the moment of inertia plays a role in determining the bat's resistance to changes in its rotational speed. A heavier bat, generally, has a larger moment of inertia, allowing it to maintain its speed through the batting zone and thus create a more forceful impact, enhancing the ball's exit velocity.
Considering the batter swinging a Wiffle ball bat, swinging at the end, furthest from the pivot point, increases the moment of inertia and thus requires more torque to achieve the same angular acceleration as grabbing the middle. However, the increased distance from the pivot allows for a higher linear speed at the end of the bat, creating a larger angular momentum on contact which could result in a more powerful impact if the batter can successfully swing the bat quickly.
In a collision, such as a bat hitting a ball, both objects exert forces on each other, as explained by Newton's third law of motion. This interaction also informs us that not only the ball is affected by the bat but the force exerted by the ball also impacts the bat, as any baseball player who has felt the sting of a ball hit off the end of the bat can attest.
A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m. a)Find the force constant of the spring. Express your answer with the appropriate units. b)The fish is now pulled down 5.00 cm and released. What is the period of oscillation of the fish? Express your answer with the appropriate units. c)What is the maximum speed it will reach? Express your answer with the appropriate units.
(a) 3539 N/m
Hook's law states that:
[tex]F=kx[/tex]
where F is the force applied on the spring, k is the spring constant, x is the stretching of the spring.
In this problem, we have:
[tex]F=mg=(65.0 kg)(9.8 m/s^2)=637 N[/tex] is the force applied (the weight of the fish)
[tex]x=0.180 m[/tex] is the stretching of the spring
Solving the equation for k, we find the spring constant:
[tex]k=\frac{F}{x}=\frac{637 N}{0.180 m}=3539 N/m[/tex]
(b) 0.85 s
The period of oscillation of a spring-mass system is
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
In this case,
m = 65.0 kg
k = 3539 N/m
Substituting into the formula,
[tex]T=2 \pi \sqrt{\frac{65.0 kg}{3539 N/m}}=0.85 s[/tex]
(c) 0.37 m/s
The initial elastic potential energy of the spring when the fish is pulled down is:
[tex]U=\frac{1}{2}k\Delta x^2[/tex]
where
[tex]\Delta x = 5.00 cm=0.05 m[/tex] is the stretching of the spring with respect to the initial position
Substituting,
[tex]U=\frac{1}{2}(3539 N/m)(0.05 m)^2=4.4 J[/tex]
The spring reaches its maximum speed when it crosses the equilibrium position, for which [tex]\Delta x=0[/tex], so when all the elastic potential energy has been converted into kinetic energy:
[tex]E=K=\frac{1}{2}mv^2[/tex]
where v is the speed of the fish. Solving for v, we find
[tex]v=\sqrt{\frac{2E}{m}}=\sqrt{\frac{2(4.4 J)}{65.0 kg}}=0.37 m/s[/tex]
The force constant of the spring is 3.53 * 10^3 N/m, the period of oscillation of the fish is 0.115 s and the maximum speed it will reach is 0.869 m/s
Explanation:A proud deep-sea fisherman hangs a 65.0-kg fish from an ideal spring having negligible mass. The fish stretches the spring 0.180 m.
By using the amount of the spring is stretched by the weight of the fish, we can calculate the force constant [tex]k[/tex] of the spring.
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex] with [tex]v_{max}=\omega A = 2 \pi f A[/tex]
When the fish hangs at rest the upward spring force [tex]|F_x| = kx[/tex] equals the weight [tex]mg[/tex]of the fish [tex]f = \frac{1}{T}[/tex]. Therefore the amplitude of the SHM is [tex]0.0500 m.[/tex]
a)Find the force constant of the spring. Express your answer with the appropriate units.
[tex]mg=kx\\k = \frac{mg}{x} \\k = \frac{65*9.8}{0.180} = 3.53 * 10^3 N/m\\[/tex]
b)The fish is now pulled down 5.00 cm and released. What is the period of oscillation of the fish? Express your answer with the appropriate units.
[tex]T = 2\pi \sqrt{\frac{m}{k}[/tex]
[tex]T = 2\pi \sqrt{\frac{65 kg}{ 3.53*10^3 N/m} [/tex]
[tex]T =0.115 s[/tex]
c)What is the maximum speed it will reach?
[tex]v_{max}=\omega A = 2 \pi f A\\v_{max}=\omega A = \frac{ 2 \pi A}{T} \\v_{max}=\omega A = \frac{ 2 \pi 0.05 m}{0.115}\\v_{max}=\omega A = 0.869 m/s[/tex]
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Suppose ocean waves are hitting a shore at a frequency of 20 waves per minute. Two swimmers are in the water. One swimmer says the frequency is 25 waves per minute and the other says the frequency is 15 waves per minute. How can the Doppler effect explain this apparent difference?
Based on the article "Will the real atomic model please stand up?," describe one major change that occurred in the development of the modern atomic model.
Answer:Which statements describe Rutherford’s model of the atom? Select all that apply.
The atom is mostly empty space.
The atom cannot be divided into smaller particles.
Electrons orbit around the center of the atom.
The atom’s positive charge is located in the atom’s nucleus.
The electrons are located within the positive material of the nucleus.
Electron clouds are regions where electrons are likely to be found.
Explanation:
Sample Response: The one major change that occurred was the placement and organization of the electron. Rutherford’s model identified that the electrons were at a distance from the nucleus, Bohr’s model identified that the electrons occurred at levels that related to their available energy, and the modern atomic model shows that electrons are located in a predicted area but cannot be identified in a specific point.
If you are testing how the amount of sleep students get affects test performance, what should you change in the experiment? A. The grading of the test B. Which teacher each student has C. How much sleep each student gets D. How many questions are on the test
C. How much sleep each student gets
It says directly that you are trying to test how the amount of sleep a student gets affects their performance. If that is the case, it would make sense to change the hours or amount of sleep each student gets before taking the test.
Suppose that the strength of the electric field about an isolated point charge has a certain value at a distance of 4 m. How will the electric field strength compare at a distance of 8 m from the point charge? 1. At twice the distance the field strength will be twice the original value. 2. At twice the distance the field strength will be 1 3 of the original value. 3. At twice the distance the field strength will be 1 2 of the original value. 4. At twice the distance the field strength will be 1 4 of the original value. 5. At twice the distance the field strength will be the same.
Answer:
4. At twice the distance the field strength will be 1/4 of the original value
Explanation:
The magnitude of the electric field produced by an isolated poitn charge is
[tex]E=k\frac{q}{r^2}[/tex]
where
k is the Coulomb's constant
q is the charge
r is the distance from the charge
In this problem, the distance from the charge is doubled from 4 m to 8 m:
[tex]r'=2 r[/tex]
Therefore, the new electric field will be
[tex]E'=k\frac{q}{(r')^2}=k\frac{q}{(2r)^2}=k\frac{q}{4r^2}=\frac{1}{4}E[/tex]
So, the field strength is 1/4 of the original value.
By doubling the distance between two charges the field strength will be [tex]\dfrac{1}{4} th[/tex] of the original value.
What will be the value of the new electric field when the distance is doubled?
The formula for the magnitude of the electric field is given by
[tex]E=k\dfrac{q}{r^2}[/tex]
Here
k is the Coulomb's constant
q is the charge
r is the distance from the charge
It is given in the problem that the distance from the charge is doubled from 4 m to 8 m:
[tex]r'=2r[/tex]
So to find the new electric field
[tex]E'=k\dfrac{q}{r^2} = k\dfrac{q}{(2r)^2} =\dfrac{1}{4} k\dfrac{q}{r^2} =\dfrac{1}{4} E[/tex]
Thus By doubling the distance between two charges the field strength will be [tex]\dfrac{1}{4} th[/tex] of the original value.
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The index of refraction for water is about 4/3. What happens to light when it travels from air into water?
When light travels from air into water, it slows down and bends towards the normal, due to the higher refractive index of water compared to air. This change in speed and direction is known as refraction. It's explained by Snell's law of refraction.
Explanation:When light travels from air into water, it notices a change in speed, due to the difference in the refractive indices of the two mediums. This change in speed causes the light to change direction, a phenomenon we know as refraction. Since the index of refraction for water (approx 1.33) is higher than that of air (approx 1.0003), the light slows down and bends towards the normal upon entering the water from the air. This is based on Snell's law of refraction, expressed as n₁ sin 0₁ = n₂ sin 0₂, where n represents the index of refraction and 0 represents the angle of incidence. Thus, the higher index of refraction in water than in air causes the beam of light to refract towards the normal.
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A series combination of two resistors, 7.25 ω and 4.03 ω, is connected to a 9.00 v battery.
a. calculate the equivalent resistance of the circuit and the current.
b. what is the potential difference across each resistor?
a. [tex]11.28\Omega[/tex]
The equivalent resistance of a series combination of two resistors is equal to the sum of the individual resistances:
[tex]R_{eq}=R_1 + R_2[/tex]
In this circuit, we have
[tex]R_1 = 7.25 \Omega\\R_2 = 4.03 \Omega[/tex]
Therefore, the equivalent resistance is
[tex]R_{eq}=7.25 \Omega + 4.03 \Omega=11.28 \Omega[/tex]
b. 5.8 V, 3.2 V
First of all, we need to determine the current flowing through each resistor, which is given by Ohm's law:
[tex]I=\frac{V}{R_{eq}}[/tex]
where V = 9.00 V and [tex]R_{eq}=11.28 \Omega[/tex]. Substituting,
[tex]I=\frac{9.00 V}{11.28 \Omega}=0.8 A[/tex]
Now we can calculate the potential difference across each resistor by using Ohm's law again:
[tex]V_1 = I R_1 = (0.8 A)(7.25 \Omega)=5.8 V[/tex]
[tex]V_2 = I R_2 = (0.8 A)(4.03 \Omega)=3.2 V[/tex]
The equivalent resistance of the circuit is 11.28 Ω, and the current through the circuit is approximately 0.798 A. The potential differences across the 7.25 Ω resistor and the 4.03 Ω resistor are approximately 5.79 V and 3.21 V, respectively.
To solve the problem, we need to brreak it down in the following steps:
Part a):
Calculate the equivalent resistance (R_eq) of the circuit:Part b):
Calculate the potential difference (V) across each resistor:Thus, the potential differences across the resistors are approximately 5.79 V and 3.21 V, respectively.
A potential difference of 12v produces a current of 0.30A in a piece of cooper wire. What is the resistance of the copper wire?
Well , its a easy problem
By ohms law,
we have V= IR
Where V is the voltage applied across the wire and R is the resistance and I is the current
=> R = V/I
=> R = 12/0.3 = 40 ohm
The resistance of the copper wire will be 400 ohms. Resistance is found as the ratio of the voltage and electric current.
What is resistance?
Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.
The given data in the problem is;
V is the voltage= 12 V
I is the current = 0.30A
R is the resistance
The resistance of the circuit from the Ohm's law is found as;
[tex]\rm V= IR \\\\ R = \frac{V}{I} \\\\\ R = \frac{12}{0.30} \\\\\ R =400 \ ohm[/tex]
Hence, the resistance of the copper wire will be 400 ohms.
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The amount of energy needed to power a 0.10-kw bulb for one minute would be just sufficient to lift a 1.0-kg object through a vertical distance of
Answer:
612.2 m
Explanation:
The work needed to lift the object is equal to its increase in gravitational potential energy:
[tex]W=\Delta U=mg \Delta h[/tex] (1)
where
m = 1.0 kg is the mass
g = 9.8 m/s^2
[tex]\Delta h[/tex] is the vertical distance
The power provided is
[tex]P=0.10 kW = 100 W[/tex]
In one minute (t = 1 min = 60 s), the work provided is
[tex]W=Pt=(100 W)(60 s)=6000 J[/tex]
Substituting this into (1) and solving for [tex]\Delta h[/tex], we find
[tex]\Delta h=\frac{W}{mg}=\frac{6000 J}{(1.0 kg)(9.8 m/s^2)}=612.2 m[/tex]
The amount of energy needed to power a 0.10-kW bulb for one minute is equivalent to the work done in lifting a 1.0 kg object through a vertical distance. The distance is calculated using the formula Work = Force * Distance. The distance is found to be 0.61 meters.
Explanation:The amount of energy needed to power a 0.10-kW bulb for one minute is 6 Joules. This energy is equivalent to the work done in lifting a 1.0 kg object through a vertical distance. To calculate the distance, we can use the formula:
Work = Force * Distance
In this case, the force is equal to the weight of the object, which is 1.0 kg * 9.8 m/s² (acceleration due to gravity). So, the distance is equal to:
Distance = Work / (Force * Acceleration due to gravity)
Substituting the values, we get:
Distance = 6 J / (1.0 kg * 9.8 m/s²) = 0.61 meters
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A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom of a gully. Its momentum upon hitting the ground is
Answer:
2940 kg m/s
Explanation:
First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:
[tex]v_y = v_0+gt[/tex]
where
v0 =0 is the initial vertical velocity
g = 9.8 m/s^2 is the acceleration due to gravity
t = 3 s is the time of the fall
Substituting into the formula, we find
[tex]v_y=0+(9.8 m/s^2)(3 s)=29.4 m/s[/tex]
And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:
[tex]p=mv=(100 kg)(29.4 m/s)=2940 kg m/s[/tex]
The motorcycle's momentum when hitting the ground is calculated using the formulas for momentum and velocity. With a mass of 100 kg and a velocity calculated as 29.4 m/s, the motorcycle's momentum is 2940 kg.m/s.
Explanation:In this case, we are asked to find the momentum of the motorcycle when it hits the ground. The momentum (p) of an object is calculated by the formula p=mv where m is the mass and v is the velocity. The velocity of the motorcycle can be calculated using the formula v=gt, where g is the acceleration due to gravity (9.8 m/s2) and t is the time. So, v=9.8*3 = 29.4 m/s. Therefore, the momentum of the motorcycle is p=100*29.4 = 2940 kg.m/s. Compared to the typical automobile with a momentum of 1400*15 = 21000 kg.m/s, the motorcycle has significantly less momentum due to its smaller mass and velocity.
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An green hoop with mass mh = 2.8 kg and radius Rh = 0.12 m hangs from a string that goes over a blue solid disk pulley with mass md = 1.9 kg and radius Rd = 0.09 m. The other end of the string is attached to a massless axel through the center of an orange sphere on a flat horizontal surface that rolls without slipping and has mass ms = 3.8 kg and radius Rs = 0.22 m. The system is released from rest. 1) What is magnitude of the linear acceleration of the hoop? m/s2 2) What is magnitude of the linear acceleration of the sphere? m/s2 3) What is the magnitude of the angular acceleration of the disk pulley? rad/s2 4) What is the magnitude of the angular acceleration of the sphere? rad/s2 5) What is the tension in the string between the sphere and disk pulley? N 6) What is the tension in the string between the hoop and disk pulley? N 7) The green hoop falls a distance d = 1.68 m. (After being released from rest.) How much time does the hoop take to fall 1.68 m?
Answer:
1) F=ma
a=(mh*g)/(7/5ms+1/2md+mh)
2) same answer as 1 because it is a system
3)a=alpha*R
alpha=a/Rd (use a from 1)
4) similar to 3
alpha=a/Rs
5)F=ma
T=ms*a
T=7/5ms*a (a from 1)
6)T=mh(g-a)
7)d=1/2at^2 (position formula - starting position and velocity are zero)
t=sqrt(2d/a) (use a from 1)
8)vf^2=vi^2 +2ad (initial velocity = 0)
vf=sqrt(2ad)
9)omega=v/Rs
Explanation:
The question involves a complex system of objects linked by a string and requires an understanding of principles like tension, angular and linear acceleration, and gravity. Using Newton's laws of motion and rotation, as well as a free-body diagram, one can identify linear and angular accelerations and string tension. The time taken for the hoop to fall can be calculated using an equation of motion.
Explanation:This is a complex system of objects attached to a string and it involves a number of different physical principles such as gravity, tension, angular acceleration, linear acceleration, and moment of inertia.
The linear acceleration of the hoop and the sphere, as well as the angular acceleration of the disk pulley and the sphere, can be identified using the Second Law of Motion (Net Force = mass x acceleration) and Newton's second law for rotation (Net Torque = Moment of Inertia x Angular Acceleration).
The tension in the string can be obtained using free-body diagrams for each object and writing equations for gravitational force and tension.
The time taken by the hoop to fall a certain distance can be calculated using the equation of motion. However, this is a complicated problem and might be better solved using numerical methods or advanced physics concepts.
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How are magnetic fields and electric fields similar
electric fields and magnetic fields both exhibit similar properties. where magnetic fields are natural electric fields are not but man made and can do the same things as a magnetic field.
A very long, straight wire carries a current of 19.0 A in the +k direction. An electron 1.9 cm from the center of the wire in the î direction is moving with a speed of 3.23 106 m/s. Find the force on the electron when it moves in the following directions. (a) directly away from the wire 0 Correct: Your answer is correct. N î + 0 Correct: Your answer is correct. N ĵ + N k (b) parallel to the wire in the direction of the current N î + 0 Correct: Your answer is correct. N ĵ + 0 Correct: Your answer is correct. N k (c) perpendicular to the wire and tangent to a circle around the wire in the + ĵ direction 0 Correct: Your answer is correct. N î + 0 Correct: Your answer is correct. N ĵ + N k
(a) [tex]1.03\cdot 10^{-16} N[/tex], -k direction
First of all, let's find the magnetic field produced by the wire at the location of the electron:
[tex]B=\frac{\mu_0 I}{2 \pi r}[/tex]
where
I = 19.0 A is the current in the wire
r = 1.9 cm = 0.019 m is the distance of the electron from the wire
Substituting,
[tex]B=\frac{(1.256\cdot 10^{-6})(19.0A)}{2 \pi (0.019 m)}=2\cdot 10^{-4} T[/tex]
and the direction is +j direction (tangent to a circle around the wire)
Now we can find the force on the electron by using:
[tex]F=qvBsin \theta[/tex]
where
[tex]q=1.6\cdot 10^{-19}C[/tex] is the electron's charge
[tex]v=3.23\cdot 10^6 m/s[/tex] is the electron speed
[tex]B=2\cdot 10^{-4} T[/tex] is the magnetic field
[tex]\theta[/tex] is the angle between the direction of v and B
In this case, the electron is travelling away from the wire, while the magnetic field lines (B) form circular paths around the wire: this means that v and B are perpendicular, so [tex]\theta=90^{\circ}, sin \theta=1[/tex]. So, the force on the electron is
[tex]F=(1.6\cdot 10^{-19}C)(3.23\cdot 10^6 m/s)(2\cdot 10^{-4} T)(1)=1.03\cdot 10^{-16} N[/tex]
The direction is given by the right hand rule:
- Index finger: direction of motion of the electron, +i direction (away from the wire)
- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)
- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means -k direction (anti-parallel to the current in the wire)
(b) [tex]1.03\cdot 10^{-16} N[/tex], +i direction
The calculation of the magnetic field and of the force on the electron are exactly identical as before. The only thing that changes this time is the direction of the force. In fact we have:
- Index finger: direction of motion of the electron, +k direction (parallel to the current in the wire)
- Middle finger: direction of magnetic field, +j direction (tangent to a circle around the wire)
- Thumb: direction of the force --> since the charge is negative, the sign must be reversed, so it means +i direction (away from the wire)
(c) 0
In this case, the electron is moving tangent to a circle around the wire, in the +j direction. But this is exactly the same direction of the magnetic field: this means that v and B are parallel, so [tex]\theta=0, sin \theta=0[/tex], therefore the force on the electron is zero.
A sample of gas with a volume of 750 ml exerts a pressure of 98 kpa at 30◦c. What pressure will the sample exert when it is compressed to 250 ml and cooled to −25◦c?
Answer:
241 kPa
Explanation:
The ideal gas law states that:
[tex]pV=nRT[/tex]
where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can rewrite the equation as
[tex]\frac{pV}{T}=nR[/tex]
For a fixed amount of gas, n is constant, so we can write
[tex]\frac{pV}{T}=const.[/tex]
Therefore, for a gas which undergoes a transformation we have
[tex]\frac{p_1 V_1}{T_1}=\frac{p_2 V_2}{T_2}[/tex]
where the labels 1 and 2 refer to the initial and final conditions of the gas.
For the sample of gas in this problem we have
[tex]p_1 = 98 kPa=9.8\cdot 10^4 Pa\\V_1 = 750 mL=0.75 L=7.5\cdot 10^{-4}m^3\\T_1 = 30^{\circ}C+273=303 K\\p_2 =?\\V_2 = 250 mL=0.25 L=2.5\cdot 10^{-4} m^3\\T_2 = -25^{\circ}C+273=248 K[/tex]
So we can solve the formula for [tex]p_2[/tex], the final pressure:
[tex]p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}=\frac{(9.8\cdot 10^4 Pa)(7.5\cdot 10^{-4} m^3)(248 K)}{(303 K)(2.5\cdot 10^{-4} m^3)}=2.41\cdot 10^5 Pa = 241 kPa[/tex]
Answer:
241 kPA
Explanation:
I just did the quest assignment, it was correct
A car engine changes chemical potential energy into the blank energy of the moving car?
Kinetic (mechanical) energy
Keesha did an experiment to study the rate of photosynthesis in the water plant Elodea. She placed a piece of Elodea in a beaker of water and set the beaker 10 centimeters (cm) from a light source. Keesha counted the bubbles released from the plant every minute for five minutes (min). She repeated the process two more times. First, she moved the light to 20 cm from the beaker, and then she moved the light to 30 cm from the beaker. What is the outcome variable (dependent variable) in this experiment? What is the key phrase that tells you what you are looking for?
What is the outcome variable in this experiment?
The dependent variable which is the outcome of bubbles produced.
Keesha did an experiment to study the rate of photosynthesis in the water plant Elodea.
My guess would be that since we initially know what the experiment is for, the key phrase is:
Keesha counted the bubbles released from the plant every minute for five minutes (min). She repeated the process two more times.
The outcome of the bubbles that are produced will be the dependent variable.
In an experiment, the variable that the researcher is testing or being measured is known as the dependent variable. For example, if a study is carried out on how tutoring has an effect on the test scores of students, in this case, the test scores of the participants will be the dependent variable.Regarding the information given in the question, the outcome of the bubbles that are produced will be the dependent variable. The outcome of the bubbles is being measured by the researcher, therefore it's the dependent variable.In conclusion, the key phrase that's vital in knowing what Keesha is looking for was "Keesha counted the bubbles released from the plant every minute for five minutes".
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Which statement describes absolute and apparent brightness?
Absolute brightness and apparent brightness are identical ways to describe the brightness of a star.
Absolute brightness and apparent brightness change with the distance from the star to the observer.
Absolute brightness depends on the distance from the star to the observer, whereas apparent brightness is always the same.
Absolute brightness is the actual amount of light produced by the star, whereas apparent brightness changes with distance from the observer.
Answer: Absolute brightness is the actual amount of light produced by the star, whereas apparent brightness changes with distance from the observer.
Explanation:
Answer: d
Explanation:
A skydiver is about to jump from an airplane that is 1,500 meters above the ground. The diver has a mass of 125 kg. What is the potential energy at this height? Show your work.
Gravitational Potential energy is the energy stored with regard to position.
the equation for potential energy is as follows
P.E = mass x gravitational acceleration x height from the ground
gravitational acceleration - 9.8 ms⁻²
substituting the values in the equation
P.E = 125 kg x 9.8 ms⁻² x 1500 m
P.E = 1 837 500 J
therefore the potential energy of the diver at 1500 m high is 1 837 kJ
The potential energy of the skydiver from the given height is 1837.5kJ.
Given the data in the question;
Mass of the diver; [tex]m = 125kg[/tex]Distance from the ground; [tex]h = 1500m[/tex]Potential energy; [tex]U = \ ?[/tex]
Gravitational potential energyGravitational potential energy is simply the potential energy an object possesses in relation to another object due to gravity.
It is expressed as:
[tex]U = mgh[/tex]
Where m is the mass of the object, h is its height from ground level and g is acceleration due to gravity ( [tex]g = 9.8m/s^2)[/tex]
Now, to determine the potential energy of the diver, we substitute our values into the expression above.
[tex]U = mgh\\\\U = 125kg\ *\ 9.8m/s^2\ *\ 1500m\\\\U = 1837500kgm^2/s^2\\\\U = 1837500J\\\\U = 1837.5kJ[/tex]
Therefore, the potential energy of the skydiver from the given height is 1837.5kJ.
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If a substance cannot be physically separated into component parts, it is likely to be
A pure substance because it cannot be separated into parts
Which of the following statements is NOT a correct assumption of the classicalmodel of an ideal gas?A. The molecules are in random motion.B. The volume of the molecules is negligible compared with the volume occupied bythe gas.C. The molecules obey Newton's laws of motion.D. The collisions between molecules are inelastic.E. The only appreciable forces on the molecules are those that occur duringcollisions.A sample of an ideal gas is in a tank of constant volume. The sample absorbsheat energy so that its temperature changes from 300 K to 600 K. If v1 is theaverage speed of the gas molecules before the absorption of heat and v2 is theiraverage speed after the absorption of heat, what is the ratio v2/ v1?A. 1/2 B. 1 C. 2 D. 2 E. 4
1) D. The collisions between molecules are inelastic
Explanation:
The kinetic theory of the gases describe the property of the gases by looking at microscopic level. At these level, some assumptions are made on the motion/collisions of the molecules of the gas:
- Molecules move by random motion
- The volume of the molecules is negligible compared with the volume of the gas
- The molecules obey Newton's laws of motion
- The intermolecular forces between the molecules are negligible except during the collisions
- Collisions between molecules are elastic
Therefore, the following statement
D. The collisions between molecules are inelastic
is wrong.
2) [tex]\sqrt{2}[/tex]
The kinetic energy Ek of a gas is directly proportional to its absolute temperature T:
[tex]E_k = \frac{3}{2}kT[/tex]
where k is the Boltzmann's constant. However, the kinetic energy depends on the square of the average velocity of the particles, [tex]v^2[/tex]:
[tex]E_k = \frac{1}{2}mv^2=\frac{3}{2}kT[/tex]
where m is the mass of the particles. This means that the velocity is proportional to the square root of the temperature:
[tex]v \propto \sqrt{T}[/tex]
So, if the temperature of the gas is doubled, the average speed increases by a factor [tex]\sqrt{2}[/tex], and the ratio v2/v1 is
[tex]\frac{v_2}{v_1}=\sqrt{2}[/tex]