A main difference between gravitational and electric forces is that electrical forces
Final answer:
The primary difference between gravitational and electric forces is that electric forces can be both attractive and repulsive, whereas gravitational forces are always attractive. Electric forces dominate on small scales, such as within atoms, while gravitational forces are significant on large scales, like between planets.
Explanation:
Main Differences Between Gravitational and Electric Forces:
A main difference between gravitational and electric forces is that electric forces can be either attractive or repulsive, while gravitational forces are always attractive. Gravitational force is proportional to the masses of the interacting objects and always pulls them together, which is important in interactions between large objects like planets and stars. Conversely, electric force is proportional to the charges of interacting objects and plays a crucial role at the microscopic level within atoms and molecules, often keeping them together. It is much stronger than gravity in most systems where both appear.
Another significant difference is that on a small scale, like that of electrons or protons, the electric force is dominant and significantly exceeds the gravitational force. In contrast, on a large scale, gravitational force typically predominates since most objects tend to be electrically neutral; thus, the net Coulomb forces cancel out.
To summarize, gravitational forces are crucial at cosmological scales, handling the structure of galaxies and the dynamics of celestial bodies, whereas electric forces are on a much smaller scale but are instrumental in the binding of atoms, the functioning of everyday materials, and biological processes.
Which statement about asteroids and meteoroids is true?
Most meteoroids orbit between Mars and Jupiter.
Asteroids frequently crash into Earth
© Meteoroids are pieces of leftover comets.
© Asteroids are pieces of meteors.
Answer:
there pieces of leftover comets :)
Explanation:
Calculate the speed of the ball, vo in m/s, just after the launch. A bowling ball of mass m = 1.5 kg is launched from a spring compressed by a distance d = 0.21 m at an angle of θ = 32° measured from the horizontal. It is observed that the ball reaches a maximum height of h = 4.4 m, measured from the initial position of the ball. Let the gravitational potential energy be zero at the initial height of the bowling ball.
Answer:
[tex]v_0=17.3m/s[/tex]
Explanation:
In this problem we have three important moments; the instant in which the ball is released (1), the instant in which the ball starts to fly freely (2) and the instant in which has its maximum height (3). From the conservation of mechanical energy, the total energy in each moment has to be the same. In (1), it is only elastic potential energy; in (2) and (3) are both gravitational potential energy and kinetic energy. Writing this and substituting by known values, we obtain:
[tex]E_1=E_2=E_3\\\\U_e_1=U_g_2+K_2=U_g_3+K_3\\\\\frac{1}{2}kd^2=mg(d\sin\theta)+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}m(v_0\cos\theta)^2[/tex]
Since we only care about the velocity [tex]v_0[/tex], we can keep only the second and third parts of the equation and solve:
[tex]mgd\sin\theta+\frac{1}{2}mv_0^2=mgh+\frac{1}{2}mv_0^2\cos^2\theta\\\\\frac{1}{2}mv_0^2(1-\cos^2\theta)=mg(h-d\sin\theta)\\\\v_0=\sqrt{\frac{2g(h-d\sin\theta)}{1-\cos^2\theta}}\\\\v_0=\sqrt{\frac{2(9.8m/s^2)(4.4m-(0.21m)\sin32\°)}{1-\cos^232\°}}\\\\v_0=17.3m/s[/tex]
So, the speed of the ball just after the launch is 17.3m/s.
Which of the following is NOT a result of supernova explosions? The neutron core is completely destroyed. Any planets within a few dozen light-years receive a life-threatening dose of radiation. Many of the elements the star fused during its life are blasted out into space. Many new elements, including some heavier than iron, are fused during the supernova explosion.
Answer:
The neutron core is completely destroyed
Explanation:
A earth - supernova is an explosion resulting to the death of a star that occurs close enough to the earth but this does not completely destroy a star. Supernovae are the most violent explosions in the universe. But they do not explode like a bomb explodes, blowing away every bit of the original bomb. Rather, when a star explodes into a supernova, its core survives. The reason for this is that the explosion is caused by a gravitational rebound effect and not by a chemical reaction. Stars are so large that the gravitational forces holding them together are strong enough to keep the nuclear reactions from blowing them apart. It is the gravitational rebound that blows apart a star in a supernova.
Final answer:
The incorrect statement is that a supernova explosion results in the complete destruction of the neutron core; often, the core becomes a neutron star. Supernovae can be life-threatening to nearby planets, but the actual danger zone is closer than a few dozen light-years. They are also responsible for creating new elements heavier than iron.
Explanation:
A supernova explosion is a cataclysmic event at the end of a massive star's life cycle. During a supernova, a number of phenomena occur, including the ejection of elements the star fused during its lifetime, and the creation of new elements heavier than iron. One of the statements provided as a result of supernova explosions is incorrect: 'The neutron core is completely destroyed'. In many cases, the core does not get completely destroyed; instead, it often collapses into a densely packed neutron star. Another result is a potential threat to life on planets within a few dozen light-years due to a life-threatening dose of radiation. It is important to note, though, that planets would need to be quite close to the supernova, much closer than a few dozen light-years, to receive such a dose. Finally, supernovae also contribute to the formation of new elements, with the violent energies enabling the fusion of atoms into elements heavier than iron.
People have proposed driving motors with the earth's magnetic field. This is possible in principle, but the small field means that unrealistically large currents are needed to produce noticeable torques. Suppose a 20-cm-diameter loop of wire is oriented for maximum torque in the earth's field.What current would it need to carry in order to experience a very modest 1.0?
Answer:
636.619772368 A
Explanation:
[tex]\tau[/tex] = Torque = [tex]1\times 10^{-3}\ N/m[/tex]
B = Magnetic field of Earth = [tex]5\times 10^{-5}\ T[/tex]
A = Area
d = Diameter = 20 cm
Current is given by
[tex]I=\dfrac{\tau}{BA}\\\Rightarrow I=\dfrac{1\times 10^{-3}}{5\times 10^{-5}\times \dfrac{\pi}{4}\times 0.2^2}\\\Rightarrow I=636.619772368\ A[/tex]
The current is 636.619772368 A
*9. A small loop of area 6.8 mm2 is placed inside a long solenoid that has 854 turns/cm and carries a sinusoidally varying current i of amplitude 1.28 A and angular frequency 212 rad/s . The central axes of the loop and solenoid coincide. What is the amplitude of the emf induced in the loop
The amplitude of the induced emf can be calculated by first finding the magnetic field within the solenoid using the formula B = µonI, calculating the magnetic flux through the coil, and then applying Faraday's law to find the time derivative of the flux. This result is then multiplied by the number of turns in the coil.
Explanation:The amplitude of the electromagnetic force (emf) induced in a coil located inside a solenoid can be calculated using Faraday's law. The first step in solving this problem involves obtaining the magnetic field B of the solenoid by using the formula B = µonI, where µo is the vacuum permeability, n represents the number of turns per unit length of the solenoid, and I is the amplitude of the current. Here, n can be obtained by converting the given number of turns per cm into turns per meter. This magnetic field B is then multiplied by the area of the coil to obtain the magnetic flux Φ through the coil.
Next, using Faraday's law, we take the time derivative of the obtained magnetic flux. Since the only variable in time here is the current I, we apply the derivative to it, with the rest of the quantities being treated as constant. Lastly, we multiply the result by the number of turns in the coil to determine the induced emf. Remember that the magnitude of the emf is the peak or 'amplitude' of the sinusoidal function associated with the time-varying current I.
https://brainly.com/question/36650405
#SPJ12
In an experiment, a large number of electrons are fired at a sample of neutral hydrogen atoms and observations are made of how the incident particles scatter. The electron in the ground state of a hydrogen atom is found to be momentarily at a distance a0/2 from the nucleus in 1 300 of the observations. In this set of trials, how many times is the atomic electron observed at a distance 2a0 from the nucleus?
Answer:
N = 1036 times
Explanation:
The radial probability density of the hydrogen ground state is given by:
[tex]p(r) = \frac{4r^{2} }{a_{0} ^{3} } e^{\frac{-2r}{a_{0} } }[/tex]
[tex]p(\frac{a_{0} }{2} ) = \frac{4(\frac{a_{0} }{2} )^{2} }{a_{0} ^{3} } e^{\frac{-2(\frac{a_{0} }{2} )}{a_{0} } }[/tex]
[tex]p(2a_{0} ) = \frac{4(2a_{0}) ^{2} }{a_{0} ^{3} } e^{\frac{-4a_{0} }{a_{0} } }[/tex]
[tex]N = 1300\frac{p(2a_{0}) }{p(\frac{a_{0} }{2} )}[/tex]
[tex]N = 1300\frac{(2a_{0}) ^{2}e^{\frac{-4a_{0} }{a_{0} } } }{(\frac{a_{0} }{2} )^{2} e^{\frac{-a_{0} }{a_{0} } }}[/tex]
[tex]N = 1300(16) e^{-3}[/tex]
N = 1035.57
N = 1036 times
. Conservation along the horizontal using a bicycle wheel: Stand on the platform holding a bicycle wheel with its axis horizontal with your arms fully extended. While holding the platform stationary, let a colleague spin the wheel. Release the platform then turn the wheel’s spin axis "up." Observe the platform turn with a ‘down’ spin vector, ω. Return the wheel to the starting position and the platform should stop. Turn the wheel’s spin axis "down." Return the wheel to the starting position. Describe the platform’s reaction in terms of conservation of vertical angular momentum: Use L1 as the angular momentum of the person and the platform, and L2 as the wheel’s angular momentum. Use vector diagrams to show how the vertical angular momentum of the system, Ltotal = L1 + L2, starts at zero and remains zero throughout the exercise. That is, show that L1i + L2i = L1f + L2f.
Answer:
w = I₂ / (I₁ -I₂) w₀ , L₂ = 2 L₁
Explanation:
This is an angular momentum exercise,
L = I w
where bold indicates vectors
We must define the system as formed by the bicycle wheel, the platform, we create a reference system with the positive sign up
At the initial moment the wheel is turning and the platform is without rotation
The initial angle moment is
Lo = L₂ = I₂ w₀
L₁ is the angular momentum of the platform and L₂ is the angular momentum of the wheel.
In the Final moment, when the wheel was turned,
[tex]L_{f}[/tex] = L₁ - L₂
L_{f} = (I₁ - I₂) w
the negative sign of the angular momentum of the wheel is because it is going downwards since the two go with the same angular velocity
as all the force are internal, and there is no friction the angular momentum is conserved,
L₀ =L_{f}
I₂ w₀ = (I₁ -I₂) w
w = I₂ / (I₁ -I₂) w₀
we can see that the system will complete more slowly
we can also equalize the angular cognition equations
L₀ = Lf
L₁ = L₂-L₁
L₂ = 2 L₁
In this part we can see that the change in the angular momentum of the platform is twice the change in the angular momentum of the wheel.
This experiment with a rotating bicycle wheel demonstrates the conservation of angular momentum. Rotating the wheel changes the platform's rotational direction to maintain a balanced angular momentum. Returning the wheel to its initial position stops the platform's rotation, preserving the system's angular momentum at zero.
In this experiment, we examine the principles of conservation of angular momentum using a rotating bicycle wheel. When you and a platform are stationary, and a colleague spins the bicycle wheel, the system's initial angular momentum is zero.
Here’s a step-by-step explanation:
Initially, both you (holding the wheel) and the platform have zero angular momentum:[tex]L_1_i + L_2_i = 0 + 0 = 0.[/tex]Your colleague spins the wheel, creating an angular momentum [tex]L_2.[/tex]If you turn the wheel's spin axis up, it creates a torque, causing you and the platform to rotate in the opposite direction. This rotation ensures that the sum of angular momenta remains zero, thus: [tex]L_1 + L_2 = 0.[/tex]Returning the wheel to the starting horizontal position stops the platform’s spin, returning the total angular momentum of the system back to zero.When you turn the wheel’s spin axis down, the platform reacts oppositely, maintaining the conservation of vertical angular momentum.After returning the wheel to horizontal, the platform stops again, demonstrating that the vertical angular momentum of the system remains balanced: [tex]L_1_f + L_2_f = 0.[/tex]This experiment vividly shows how changing the direction of angular momentum through torque affects rotational motion while conserving the overall angular momentum of the system.
The angular speed of an automobile engine is increased at a constant rate from 1150 rev/min to 2680 rev/min in 14.2 s. (a) What is its angular acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this 14.2 s interval?
Answer:
(A) [tex]\alpha =11.277rad/sec^2[/tex]
(B) [tex]\Theta =2846.30rad[/tex]
Explanation:
We have given initial angular speed
[tex]\omega _i=1150rpm=\frac{2\times 3.14\times 1150}{60}=120.366rad/sec[/tex]
[tex]\omega _f=2680rpm=\frac{2\times 3.14\times 2680}{60}=280.506rad/sec[/tex]
Time t = 14.2 sec
(a) From first equation of motion
[tex]\omega _f=\omega _i+\alpha t[/tex]
[tex]280.506=120.366+\alpha \times 14.2[/tex]
[tex]\alpha =11.277rad/sec^2[/tex]
(b) From third equation of motion
[tex]\omega _f^2=\omega _i^2+2\alpha \Theta[/tex]
[tex]280.506^2=120.366^2+2\times 11.277\times \Theta[/tex]
[tex]\Theta =2846.30rad[/tex]
What are the three bone "shelves" inside the nasal cavity called? *
Vour ancier
Answer: the nasal conchae also known as the turbinates
Explanation:
List down different layers of the sun. Rank these layers based on their distance from the sun’s center
Answer:
Core
Radiative zone
Convective zone
Photosphere
Chromosphere
Transient region
Corona
Ranks of layers based on their distance from the sun’s center
1st-corona
2nd-Transient region
3rd-chromosphere
4th-Photosphere
5th-convective zone
6th-radiative zone
7th-core
A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m.What is the magnitude ofthe normal force exerted on the car by the walls of the circle at pointA?
Answer:
61.76 N.
Explanation:
Given the mass of the car, m = 1.60 kg.
The speed of the car, v = 12.0 m/s.
The radius of the circle, r = 5 m.
As car is moving in circular motion, so net force ( normal force + weight of the car) is equal to centripetal force enables the car to reamins in circular path.
Let N is the normal force.
So, [tex]N - mg = F_c[/tex]
[tex]N-mg=\frac{mv^2}{r}[/tex]
Now substitute the given values, we get
[tex]N-1.60kg\times9.8m/s^2=\frac{1.60kg\times(12.0m/s)^2}{5.0m}[/tex]
[tex]N=15.68+46.08[/tex]
N = 61.76 N.
Thus, the magnitude ofthe normal force exerted on the car by the walls is 61.76 N.
The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In this case, the magnitude of the normal force is equal to the weight of the car, which is the product of its mass and the acceleration due to gravity.
Explanation:The magnitude of the normal force exerted on the car by the walls of the circle at point A can be calculated using the concept of centripetal force. In a vertical circle, the net external force equals the necessary centripetal force. The only two external forces acting on the car are its weight and the normal force of the road. Since the car is not leaving the surface and the net vertical force must be zero, the vertical components of the two external forces must balance each other.
The vertical component of the normal force is equal to the car's weight, which is mg. The weight can be calculated by multiplying the mass of the car by the acceleration due to gravity. In this case, the weight is 1.60 kg multiplied by 9.8 m/s². Therefore, the magnitude of the normal force exerted on the car at point A is also 1.60 kg multiplied by 9.8 m/s².
Learn more about Centripetal Force here:https://brainly.com/question/11324711
#SPJ3
Each of the following would cause an increase in blood pressure except __________.
Answer:
an inhibitor of angiotensin II
Explanation:
Angiotensin, specifically angiotensin II binds to many receptors in the body to affect several systems. It can normally increase blood pressure by constricting the blood vessels but with the introduction of an inhibitor, it wouldn't bring about an increase in blood pressure.
. A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00m above the ground. At the same instant the balloon is released; a second balloon is thrown straight down at 2.00 m/s from the same height. Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon
Answer:
Explanation:
Height of building
H = 6m
Horizontal speed of first balloon
U1x = 2m/s
Second ballot is thrown straight downward at a speed of
U2y = 2m/s
Time each gallon hits the ground
Balloon 1.
Using equation of free fall
H = Uoy•t + ½gt²
Uox = 0 since the body does not have vertical component of velocity
6 = ½ × 9.8t²
6 = 4.9t²
t² = 6 / 4.9
t² = 1.224
t = √1.224
t = 1.11 seconds
For second balloon
H = Uoy•t + ½gt²
6 = 2t + ½ × 9.8t²
6 = 2t + 4.9t²
4.9t² + 2t —6 = 0
Using formula method to solve the quadratic equation
Check attachment
From the solution we see that,
t = 0.9211 and t = -1.329
We will discard the negative value of time since time can't be negative here
So the second balloon get to the ground after t ≈ 0.92 seconds
Conclusion
The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.
Answer:
Second balloon hits ground Δt = 0.185 seconds sooner than first balloon
Explanation:
Given:-
- The first balloon is thrown horizontally with speed, u1 = 2.0 m/s
- The second balloon is thrown down with speed, u2 = 2.0 m/s
- The height from which balloon are thrown, si = 6.0 m (above ground)
Find:-
Determine which balloon hits the ground first and how much sooner it hits the ground than the other balloon
Solution:-
- We will first determine the time taken (t1) for the first balloon thrown horizontally with speed u1 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.
- Using the second kinematic equation of motion in vertical direction:
si = 0.5*g*t1^2
Where, g: The gravitational constant = 9.81 m/s^2
6.0 = 4.905*t1^2
4.905*t1^2 - 6.0 = 0
- Solve the quadratic equation:
t 1 = 1.106 s
- Similarly, the time taken (t2) for the second balloon thrown down with speed u2 = 2.0 m/s from top of building from a height of s = 6.0 m from ground to it the ground.
- Using the second kinematic equation of motion in vertical direction:
si = u2*t2 + 0.5*g*t1^2
Where, g: The gravitational constant = 9.81 m/s^2
6.0 = 4.905*t1^2 + 2*t2
4.905*t1^2 + 2*t2 - 6.0 = 0
- Solve the quadratic equation:
t 2 = 0.9208 s
- We see that the second balloon thrown down vertically hits the ground first. The second balloon reaches ground, t1 - t2 = 0.185 seconds, sooner than first balloon.
Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?
Answer:
15 miles
Explanation:
6 miles per hour
2 1/2 hours
6 x 2 = 12
6 x 1/2 = 3
12 + 3 = 15
Two rods are made of brass and have the same length. The cross section of one of the rods is circular, with a diameter of 2a. The other rod has a square cross section, where each side of the square is a length 2a. One end of the rods is attached to an immovable fixture which allows the rods to hang vertically. To the free end of each rod, a block of mass m is attached. Which rod, if either, will stretch more after the block is attached?
A. The one with the circular cross section will stretch more.
B. The one with the square cross section will stretch more.
C. Both will stretch by the same amount.
D. One cannot say which will stretch more without knowing the numerical values of a and m.
Answer:
A. The one with the circular cross section will stretch more.
Explanation:
According to the given data:
Two rods are made of brass and have the same length
Both rods having circular and square cross-section
Diameter of circular cross-section given is 2 a
therefore, Cross-section = [tex]A_c=\frac{\pi (2a)^2}{4}=\pi a^2[/tex]
If the length of square=2 a
then, Cross-section = [tex]A_{s}[/tex] = (2a)²=>4a²
Change in Length of rod = PL / AE
δL[tex]\alpha \frac{1}{A}[/tex]
Now, we are considering other factors same
the area of cross-section of square rod is more than Area of cross-section of circular rod
thus, the one with the circular cross section will stretch more
Answer:
The correct option is;
A. The one with the circular cross section will stretch more.
Explanation:
Here we have the cross section as being
1. Circular, with diameter, D = 2·a
2. Square cross section with each side length = 2·a
The area of the circular rod is then
Area of circle = π·D²/4 which is equal to
π×(2·a)²/4 = π·4·a²/4 = π·a²
The area of the rod with square cross section is
Area of square = Side² which gives
Area of cross section = (2·a)² = 4·a²
Therefore, since π = 3.142, the cross sectional area of the circular rod is less than that of the one with a square cross section
That is, π·a² = 3.142·a² < 4·a²
We note that the elongation or extension is directly proportional to the force applied as shown as follows
[tex]\frac{P}{A} = E\frac{\delta}{L}[/tex]
Where:
P/A = Force and
δ = Extension
The force is inversely proportional to the area, therefore a rod with less cross sectional area experiences more force and more elongation.
The gravitational potential energy of an object is equal to its weight multiplied by its
Answer:
Height above a surface
Explanation:
Gravitational potential energy is the energy which an object possesses due to its position above a surface.
It is also the amount of work a force has to do in order to bring an object from a particular position to a point of reference.
It is given mathematically as:
P. E. = m*g*h
where m = mass of the body
g = acceleration due to gravity
h = height above a surface
m*g represents the weight of the object.
Hence, Gravitational potential energy is the product of an object's weight and its height above a surface/reference point.
Each of the two wires in a particular 16-gauge extension cord has an electrical resistance of 0.04 V. You're using this extension cord to operate a toaster oven, so a current of 15 A is fl owing through it. What is the voltage drop across each wire in this extension cord?
The voltage drop across each wire in the extension cord with a resistance of 0.04 ohms and a current of 15 A flowing through it is calculated using Ohm's Law (V = I x R) to be 0.6 volts.
To calculate the voltage drop across each wire in the extension cord, we can apply Ohm's Law, which states that voltage (V) is the product of current (I) and resistance (R). Given that each wire has a resistance of 0.04 Ω and a current of 15 A is flowing through it, we can calculate the voltage drop by multiplying these two figures:
V = I × R
V = 15 A × 0.04 Ω = 0.6 V
Therefore, the voltage drop across each wire in the extension cord is 0.6 volts.
Which element is a metal? A.lithium B.Nitrogen C.sulfur D.chlorine
Answer:
the answer is b
Explanation:
hope it helps ...
Lithium (Li) is a metal located in Group 1 of the periodic table and is known as an alkali metal. It stands out for its low density and capacity to conduct electricity, along with exhibiting the typical properties of metals.
The element lithium (Li) is a metal, and notably, it is the first metal in the periodic table located in Group 1. It has the lowest atomic number (3) among the metals. The defining characteristics of metals include their ability to conduct electricity, possess malleability, and often exhibit a shiny luster. Lithium, being part of the alkali metals category, shares these properties. Furthermore, lithium is the least dense metal and has a unique electronic structure where the third electron occupies an outer shell, differentiating it from nonmetals.
Lithium is used in various applications including batteries and the aerospace industry due to its low density and high reactivity. The chemically similar elements to lithium include other alkali metals such as sodium and potassium, which also share similar chemical properties.
A cue ball of mass m1 = 0.34 kg is shot at another billiard ball, with mass m2 = 0.575 kg, which is at rest. The cue ball has an initial speed of v = 7.5 m/s in the positive direction. Assume that the collision is elastic and exactly head-on.
a) write an expression for the horizontal component of the billiard ball's velocity, vr after the collision, in terms of the other variables of the problem.
b) what is this velocity, in meters per second?
c) Write an expression for the horizontal component of the cue ball's velocity, vr, after the collision.
d) what is the horizontal component of the cue ball's final velocity, in meters per second?
Answer:
Part(a): The expression for the velocity of the billiard ball is [tex]\bf{v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}}[/tex]
Part(b): The value of the velocity of the billiard ball is [tex]\bf{5.57~m/s}[/tex].
Part(c): The expression for the velocity of the cue ball is [tex]\bf{v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}}[/tex]
Part(d): The value of the velocity of the cue ball is [tex]\bf{1.93~m/s}[/tex].
Explanation:
Given:
The mass of the cue ball, [tex]m_{1} = 0.34~kg[/tex].
The mass of the billiard ball, [tex]m_{2} = 0.575~kg[/tex].
The initial velocity of the cue ball, [tex]u_{1} = 7.5~m/s[/tex]
The initial velocity of the billiard ball, [tex]u_{2} = 0[/tex]
(a)
Consider the final velocity of the cue ball be [tex]v_{1}[/tex] and the final velocity of the billiard ball be [tex]v_{2}[/tex].
From the conservation of linear momentum , we can write
[tex]~~~~&& m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}\\&or,& m_{1}(u_{1} - v_{1}) = m_{2}(v_{2} - u_{2})~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
From the conservation of energy, we can write
[tex]~~~&& \dfrac{1}{2}m_{1}u_{1}^{2} + \dfrac{1}{2}m_{2}u_{2}^{2} = \dfrac{1}{2}m_{1}v_{1}^{2} + \dfrac{1}{2}m_{2}v_{2}^{2}\\&or,& \dfrac{1}{2}m_{1}(u_{1}^{2} - v_{1}^{2}) = \dfrac{1}{2}m_{2}(v_{2}^{2} - u_{2}^{2})~~~~~~~~~~~~~~~~~~~~~~~(2)[/tex]
Dividing equation (1) by equation (2), we have
[tex]~~~&& \dfrac{m_{1}(u_{1} - v_{1}) }{m_{1}(u_{1}^{2} - v_{1}^{2})} = \dfrac{m_{2}(v_{2} - u_{2})}{m_{2}(v_{2}^{2} - u_{2}^{2})}\\&or,& u_{1} + v_{1} = u_{2} + v_{2}~~~~~~~~~~~~~~~~~~~~~~~~(3)[/tex]
Rearranging equation (3) for [tex]v_{1}[/tex], we have
[tex]v_{1} = u_{2} + v_{2} - u_{1}~~~~~~~~~~~~~~~~~~~~(4)[/tex]
Substitute equation (4) in equation (1), we can write
[tex]v_{2} = \dfrac{2m_{1}u_{1}}{m_{1}+m_{2}} + \dfrac{m_{2} - m_{1}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(5)[/tex]
Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{2f}[/tex] for [tex]v_{2}[/tex] in equation (5), we have
[tex]v_{2f} = \dfrac{2m_{1}v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(6)[/tex]
(b)
Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (6), we have
[tex]v_{2f} &=& \dfrac{2(0.34~kg)(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& 5.57~m/s[/tex]
(c)
Rearranging equation(3) for [tex]v_{2}[/tex], we have
[tex]v_{2} = u_{1} + v_{1} – u_{2}~~~~~~~~~~~~~~~~~~~~(7)[/tex]
Substitute equation (7) in equation (1), we can write
[tex]v_{1} = \dfrac{2m_{2}u_{2}}{m_{1}+m_{2}} + \dfrac{m_{1} - m_{2}}{m_{1} + m_{2}}u_{2}~~~~~~~~~~~~~~(8)[/tex]
Substituting [tex]0[/tex] for [tex]u_{2}[/tex], [tex]v[/tex] for [tex]u_{1}[/tex] and [tex]v_{1f}[/tex] for [tex]v_{1}[/tex] in equation (8), we have
[tex]v_{1f} = \dfrac{(m_{1} - m_{2})v}{m_{1}+m_{2}}~~~~~~~~~~~~~~~~~~~~~~~~(9)[/tex]
(d)
Substituting [tex]0.34~kg[/tex] for [tex]m_{1}[/tex], [tex]7.5~m/s[/tex] for [tex]v[/tex] and [tex]0.575~kg[/tex] for [tex]m_{2}[/tex] in equation (9), we have
[tex]v_{1f} &=& \dfrac{(0.34 - 0.575)~kg(7.5~m/s)}{(0.34 + 0.575)~kg}\\~~~~~&=& -1.93~m/s[/tex]
Negative sign indicates that the cue ball will bounce back.
The expressions for the horizontal component of the billiard ball's velocity after the collision and the cue ball's velocity after the collision are given. The velocities can be calculated using the provided equations and given values.
Explanation:a) The expression for the horizontal component of the billiard ball's velocity, vr after the collision is given by:
vr = (m1 - m2) * (v1 / (m1 + m2))
b) Substituting the given values into the equation, we find that the horizontal component of the billiard ball's velocity after the collision is approximately 4.02 m/s.
c) The expression for the horizontal component of the cue ball's velocity, vr after the collision is given by:
vr = (2 * m2 * v1) / (m1 + m2)
d) Substituting the given values into the equation, we find that the horizontal component of the cue ball's final velocity is approximately 2.48 m/s.
Learn more about Billiard ball collision here:https://brainly.com/question/15152823
#SPJ3
We all depend on electricity. Most electricity is created by electromagnetic generators at large power plants and distributed through an electric grid. The electric grid in the United States is interconnected and includes more than 400,000 miles of electric transmission lines. However, the grid has not been well maintained, and the number and severity of power outages has been rising over the past decade. Because the grid is interconnected, a severe power outage in one area can quickly spread to other areas.
How long could you survive without electricity? What parts of your life would be affected by loss of electricity? Should you prepare for an electricity outage, and if so, how would you prepare? What backup system could your family or community install to generate limited amounts of electricity during an outage? How does this system create an electric force field and generate electric current?
Answer:
1) not so long (maybe an hour or two)
2) access to information through the internet will be most affected if my computer and mobile phone run out of battery power.
3) yes, one should prepare for power outage. This can be done by having a standby alternative source of power like the use of inverters that stores electrical energy in form of chemical energy, and small internal combustion engine powered electric generators.
4) solar panels can be used to draw power from incident sun rays, this power can be stored in an inverter for future use in case of a power outage.
5) energy from the sun is converted into direct current which is then supplied to an accumulator in the opposite direction to its flow of current. When the energy is needed, it can be used directly, or converted to an alternating current. This is achieved by connecting its terminal to the supply. Electric field is generated by flow of ions and electrons within the working chemical (e.g lithium).
Explanation:
Answer:
If there was a power outage, you could survive normally using older ways of life and with proper preparations; some ways to prepare is having canned foods that don't require heating or refrigeration, since you wouldn't have electronic means of preparing it, and supplies for starting camp fires. Some ways your life would be affected is lack of simple methods of day-to-day tasks that you have become accustomed to: no warm water, to refrigerators, or heating systems such as electronic kettles, microwaves or ovens. A backup system you and/or your family/community could install would be solar panels, to collect power during the day to then be used at night or when needed. Solar panels collect solar energy by catching the sun-rays in the cells of the panels and turning the energy into a direct current (DC), then the panels convert the direct current into usable alternating current (AC) energy with the help of inverter technology. That current is then distributed throughout the building accordingly.
A horizontal vinyl record of mass 0.115 kg and radius 0.0896 m rotates freely about a vertical axis through its center with an angular speed of 5.58 rad/s and a rotational inertia of 4.84 x 10-4 kg·m2. Putty of mass 0.0484 kg drops vertically onto the record from above and sticks to the edge of the record.What is the angular speed of the record immediately afterwards?
Answer:
Explanation:
Angular momentum is the product of inertial and angular frequency
L = I × ω
Where
L is angular momentum
I is inertia
ω is angular frequency
So, given that
Vinyl record has a massz
M = 0.115kg
Radius R = 0.0896m
Angular velocity of vinyl record
ω(initial) = 5.58 rad/s
Rotational inertial of vinyl record.
I(initial) = 4.84 × 10^-4 kgm²
Putty drop on the record
Mass of putty M' = 0.0484kg
Angular speed after putty drop ω'
Using conversation of angular momentum
Initial angular momentum is equal to final angular momentum
I(initial) × ω(initial) = I(final) × ω(final)
So, we need to find I(final)
Inertia log putty can be determine using MR² by assuming a thin loop
I(putty) = M'R² = 0.0484 × 0.0896
I(putty) = 3.89 × 10^-4 kgm²
I(final) = I(initail) + I(putty)
I(final) = 4.84 × 10^-4 + 3.89 × 10^-4
I(final) = 8.73 × 10^-4 kgm²
Therefore,
I(initial) × ω(initial) = I(final) × ω(final)
ω(final) = I(initial) × ω(initial) / I(final)
ω(final) = 4.84 × 10^-4 × 5.58 / 8.73 × 10^-4
ω(final) = 3.1 rad/s
An open container holds ice of mass 0.550 kg at a temperature of -15.3 âC . The mass of the container can be ignored. Heat is supplied to the container at the constant rate of 900 J/minute .The specific heat of ice to is 2100 J/kgâK and the heat of fusion for ice is 334Ã103J/kg.a) How much time tmelts passes before the ice starts to melt?tmelts=___minutesb)From the time when the heating begins, how much time trise does it take before the temperature begins to rise above 0âC?trise=____minutes
The ice in the open container begins to melt after 19.82 minutes of heating, and the temperature of the system starts to rise above 0°C after a total heating time of 223.93 minutes.
Explanation:To answer both parts of this question, we need to calculate the time for two processes: the heating of the ice to 0°C (melting point), and the melting of the ice into water at 0°C.
Firstly, we find the heat (Q) required to raise the temperature of the ice to 0°C using Q=mcΔT, which is 0.55kg * 2100 J/kg°C * (0 - (-15.3°C)) = 17842.5 J. As the heater adds 900J per minute, the time for this is 17842.5J ÷ 900J/min = 19.82 minutes.
Next, we find the heat required to melt ice into water at 0°C using Q=mLf, where Lf = 334000 J/kg. This is 0.55kg * 334000 J/kg = 183700J. The time for this can be found by 183700J ÷ 900J/min = 204.11 minutes.
tmelts (the time before ice begins to melt) is 19.82 minutes, and trise (the total time before the temperature begins to rise above 0°C) is 19.82 min + 204.11 min = 223.93 minutes.
Learn more about Heat Transfer Calculations here:https://brainly.com/question/31080599
#SPJ12
The time before the ice starts to melt (tmelts) is 19.63 minutes, and the total time before the temperature begins to rise above 0 °C (trise) is 223.74 minutes.
Explanation:Calculating Time Until Ice Starts to MeltTo calculate the time tmelts before the ice starts to melt, you firstly need to determine how much heat is necessary to raise the ice from -15.3 °C to 0 °C using the formula Q = m * c * ΔT, where Q is the heat, m is the mass of the ice, c is the specific heat of ice, and ΔT is the change in temperature. In this case, Q would be:
Q = (0.550 kg) * (2,100 J/kg·K) * (15.3 K) = 17,661.5 J.
Since heat is supplied at 900 J/minute, the time tmelts can be calculated as:
tmelts = Q / (heat rate) = 17,661.5 J / (900 J/min) = 19.63 minutes.
Calculating Time Until Temperature Rises Above 0 °CTo calculate the time trise until the temperature begins to rise above 0 °C, we must add the time it takes to melt the ice completely at 0 °C. The heat of fusion (∆Hfus) formula is used here: Q = m * Lf where Lf is the heat of fusion for ice. Considering that the heat of fusion for ice is 334,000 J/kg:
Q = (0.550 kg) * (334,000 J/kg) = 183,700 J.
For melting the ice at 0 °C, it will take:
trise = Q / (heat rate) = 183,700 J / (900 J/min) = 204.11 minutes.
Therefore, the total time before the temperature begins to rise above 0 °C is the sum of the time to warm the ice to 0 °C and the time to melt it completely, which will be 19.63 minutes + 204.11 minutes = 223.74 minutes.
Learn more about Heat Transfer in Ice here:https://brainly.com/question/18940221
#SPJ3
Shannon is playing tennis. She swings her racket to hit the tennis ball and sends the ball flying in the same direction.
Answer:
she used force to swing her ball all way back
Explanation:
Final answer:
Shannon swinging her racket to hit a tennis ball illustrates key physics concepts, including momentum and energy transfer, crucial for propelling the ball forward. The energy transferred from the racket to the ball depends on the racket's motion, enhancing the ball's speed when the racket moves toward it. Physics principles, such as Newtonian mechanics, are essential for understanding and improving tennis skills.
Explanation:
When Shannon swings her racket to hit a tennis ball, sending it flying in the same direction, several principles of physics come into play, particularly related to momentum and energy transfer. First, when the racket contacts the ball, momentum is transferred from the racket to the ball, propelling it forward. According to Newton's third law, for every action, there is an equal and opposite reaction. Therefore, as the racket exerts force on the ball, the ball exerts an equal and opposite force on the racket, though the effects are more noticeable on the ball due to its much smaller mass.
Energy transfer during the collision between the racket and the ball is significant. A moving racket transfers kinetic energy to the stationary ball, which then moves with a velocity dependent on the imparted energy. If the racket is moving towards the ball, the speed of the ball post-collision is greatly increased. This results from the addition of the racket's velocity to that imparted to the ball, showcasing an efficient energy transfer.
Understanding the dynamics of hitting a tennis ball, including the effects of the racket's speed, angle of swing, and point of impact, requires knowledge of Newtonian mechanics and is essential for improving one's tennis playing skills. Moreover, accounting for the arc of the ball and its spin are crucial for predicting how the ball will move, enhancing the player's ability to plan their shots strategically.
The energy required to initiate an exergonic reaction is called
Answer:
energy of activation is the correct answer.
Explanation:
All the chemical reactions including an exothermic reaction require activation energy to initiate a reaction.Activation energy is the minimum amount of energy required to activate the molecules to the situation that they can experience a chemical change. If the activation energy is lower the faster the reaction occurs.Joules, kilocalories/mol, kilojoules per mole are the units for activation energy.The activation energy is the energy needed to initiate an exergonic reaction. It is required to kickstart the process, despite the reaction being one that eventually releases energy.
Explanation:The energy required to initiate an exergonic reaction is called the activation energy. This is the minimum amount of energy required for a chemical reaction to occur. In an exergonic reaction, the activation energy is critical even though these reactions release energy overall, because it's needed to start the process. To envision this, think of pushing a boulder off a hill; the initial push (activation energy) is required for the action (the exergonic reaction) to begin, after which gravity (the reaction's natural tendency) takes over and the boulder rolls down (energy is released).
Learn more about Activation Energyhttps://brainly.com/question/28384644
#SPJ6
Two skaters, one with mass 65 kg and the other with mass 40 kg, stand on an ice rink holding a pole of length 10 m and negligible mass. Starting from the ends of the pole, the skaters pull themselves along the pole until they meet. How far does the 40 kg skater move
The 40 kg skater moves approximately 6.19 meters towards the 65 kg skater when they pull themselves along a 10 m pole on ice, utilizing the conservation of momentum and considering the center of mass of the system.
Explanation:To solve how far the 40 kg skater moves when two skaters pull themselves along a pole on ice, we use the principle of conservation of momentum. This principle states that in the absence of external forces, the total momentum of a system remains constant. Since both skaters start from rest, their initial total momentum is zero. As they pull towards each other, their momenta are equal in magnitude and opposite in direction, keeping the total momentum of the system at zero.
Let's denote the distance the 65 kg skater moves as d65 and the distance the 40 kg skater moves as d40. Because the pole has a length of 10 m, these two distances must add up to 10 m, so d65 + d40 = 10. To find out how far each skater moves, we look at their center of mass. The system's center of mass doesn't change because there are no external forces acting on the system.
The position of the center of mass xCM can be found using the formula xCM = (m1 * x1 + m2 * x2) / (m1 + m2), where m1 and m2 are the masses of the skaters, and x1 and x2 are their respective positions. Considering the system initially (before movement), the center of mass is at the midpoint of the pole as the skaters are at the ends. Therefore, xCM = 5 m.
Because the total mass of the system is 105 kg and the center of mass is 5 m from the starting point of either skater, we find that the 40 kg skater moves closer to the center of mass than the 65 kg skater to maintain the center of mass position. By directly calculating it, d40 = 10 * (65 / 105) = 6.19 m. Thus, the 40 kg skater moves approximately 6.19 meters towards the 65 kg skater.
A chain lying on the ground is 11 meters long and its mass is 95 kilograms. The chain is threaded through a pulley, which is fixed to the ground, and pulled directly up so that it forms the shape of an L. How much work is required to raise one end of the chain to a height of 7 meters?
Answer: 296.1 J or 6.98 kJ
Explanation:
Given
Length of chain, l = 11 m
Mass of chain, m = 95 kg
It is worthy of note that chain has two ends. So this depends which of the two ends should end up at
the given height of 7 m.
Scenario 1. If it is the leading end:
Weight of chain raised
W = mg/l
W = 9.8×95/11
W = 84.6 N
Height raised = ½ m (which is the height of the centre of gravity of the raised portion).
Work done = 84.6 × 1/2 * 7 = 296.1 Joules
Scenario 2. If it is the trailing end:
Weight of chain raised = 9.8×95 = 931 N.
Height raised (average) = 12.5 m (5.5 m for the chain midpoint + 7 m off the ground)
Work done = 931 × 7.5 = 6.98 kJoules
Thus, the work required is either of 296.1 J or 6.98 kJ
The work required to lift one end of a 95 kg chain to 7 meters height is 3261.85 J.
To calculate the work required to raise one end of the chain to a height of 7 meters, we consider the work done against the force of gravity. The chain has a mass of 95 kilograms, and we need to move it through a vertical distance of 7 meters.
Work is defined as the product of force and the distance over which the force is applied. In this case, the force is the weight of the chain, which is the product of its mass and the acceleration due to gravity (g = 9.81 m/s²). However, the chain gets lighter as you pull it up, since a smaller length of chain is left hanging at every moment. To solve this, we find the average force, which would be the weight of half the chain, as only half of it will be lifted through the full height on average.
The chain's total weight is mass multiplied by the acceleration due to gravity so we have 95 kg × 9.81 m/s².
The weight of half the chain is therefore
0.5 × 95 kg × 9.81 m/s²
The work done to lift half the mass through the full height is the average force times the height (7 m).
The calculation will then be:
{W} = 0.5 ×95 kg × 9.81 m/s² ×7 m
W= 3261.85 J
So, we multiply these values to find the total work done in lifting one end of the chain to a height of 7 meters.
A high-speed flywheel in a motor is spinning at 450 rpm (revolutions per minute) when a power failure suddenly occurs. The flywheel has mass 45.0 kg and diameter 74.0 cm . The power is off for 25.0 s and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 180 complete revolutions. A. At what rate is the flywheel spinning when the power comes back on? (rad/s)B. How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on? (s)C. How many revolutions would the wheel have made during this time?
Answer:
a) final angular speed w2 = 414rpm
b) time to stop t = 312.5s
c) angular distance from start to stop s = s = 1171.88 rev
Explanation:
Average angular speed wa = (w1+w2)/2 = Angular distance/time = s/t
Given;
w1 = 450 rpm = 450/60 rev/s = 7.5 rev/s
s = 180 rev
t = 25s
a)
s/t = (w1+w2)/2
w2 = 2s/t - w1
Substituting the values;
w2 = 2×180/25 - 7.5 = 6.9 rev/s
w2 = 6.9×60rpm = 414rpm
b)
Angular deceleration a = (w1-w2)/t = (7.5-6.9)/25
a = 0.024 rev/s^2
time t for w2 = 0
t = (w1-w2)/a = (7.5-0)/0.024
t = 312.5 s
c)
Angular distance from start to stop s;
s = w1t - 0.5at^2
s = 7.5×312.5s - 0.5×0.024×312.5^2
s = 1171.88 rev
Si te dijera que la energía es una cualidad o capacidad que tienen los cuerpos materiales ¿para qué sirve o emplean los cuerpos esa cualidad o capacidad?
Answer:
La energía es una cualidad o capacidad que tienen los cuerpos materiales y posee diversas funciones. Hay distintos tipos de energía, pero ésta permite que un cuerpo se mueva por ejemplo, cambie de temperatura o de estado.
Explanation:
La Energía de los cuerpos puede ser definida de distintas maneras según la ciencia en la cual estemos trabajando. Para la física, la energía es la capacidad para realizar un trabajo. Esa energía puede ser cinética, térmica, química, y cada una de ellas define distintas propiedades de los cuerpos, que hacen que puedan cambiar de posición, temperatura, estado.
You sit at the middle of a large turntable at an amusement park as it is set spinning on nearly frictionless bearings, and then allowed to spin freely. When you crawl toward the edge of the turntable, does the rate of the rotation increase, decrease, or remain unchanged, and why
Answer:
Decrease
Explanation:
If you crawl to the rim the rotational speed will decrease. The law of conservation of angular momentum supports this answer. And it states that :
"When the net external torque acting on a system about a given axis is. zero , the total angular momentum of the system about that axis remains constant."
Final answer:
When moving toward the edge of a spinning turntable, the rate of rotation decreases due to the principle of conservation of angular momentum, which necessitates a decrease in angular velocity to compensate for an increased moment of inertia.
Explanation:
When you crawl toward the edge of a large turntable at an amusement park while it is spinning, the rate of the rotation decreases. This phenomenon is explained by the principle of conservation of angular momentum, which states that if no external torque acts on a system, the total angular momentum of the system remains constant.
Angular momentum is given by the product of the moment of inertia (I) and the angular velocity (ω), represented by the equation L = Iω. As you move away from the axis of rotation, your moment of inertia increases because the moment of inertia is directly proportional to the square of the distance from the axis of rotation. In order to conserve angular momentum, if the moment of inertia increases, the angular velocity must decrease accordingly.