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A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answers

Answer 1

Answer:

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

$V_f^2=V_o^2-2gy$

As the spheres are released v₀ = 0

$V_f^2=2gH$

$V_f=√(2gH)$

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

$V_(10)=\sqrt2gH$

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

$V_h=\sqrt2gH$

Small sphere m₂ and $V_(20)=-\sqrt2gH=-V_h$

Large sphere m₁ and $V_(10)=√(2gH)=V_h$

Before crash

$P_o=m_1V_(10)+m_2V_(20)$

After the crash

$P_f=m_1V_(1f)+m_2V_(2f)$

$P_o=P_f$

$m_1V_(10)+m_2V_(20)=m_1V_(1f)+m_2V_(2f)$

The conservation of kinetic energy

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$

$K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

$K_o=K_f$

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$ $=$ $K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

Let's write the values

$-m_1V_h+m_2V_h=m_1V_(1f)+m_2V_(2f)$

$m_1V_h^2+m_2V_h^2=m_1V_(1f)^2+m_2V_(2f)^2$

The solution to this system of equations is

$m_t=m_1+m_2$

$V_(1f)=((m_1-m_2))/(m_tV_(10)) +(2m_2)/(m_tV_2)$

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_2)$

The large sphere is labeled 1, we are asked for the mass so that $V_(1f)$ = 0, let's clear the equation

$V_(1f)=(m_1-m_2)/(m_tV_(10)) +(2m_2)/(m_tV_(20))$

$0=(m_1-m_2)/(m_tV_h) +(2m_2)/(m_t(-V_h))$

$(m_1-m_2)/(m_tV_h) =(2m_2)/(m_tV_h)$

$(m_1-m_2)=2m_2$

$m_1=3m_2$

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_(20))$

$V_(2f)=(2m_1)/(m_tV_h) +(m_2-m_1)/(m_t(-V_h))$

In addition, we know that m₁ = 3 m₂

$m_t=3m_2+m_2$ mt = 3m2 + m2

$m_t=4m_2$

$V_(2f)=(2* 3m_2)/(4m_2V_h-(m_2-3m_2)4m_2V_h)$

$V_(2f)=(3)/(2) V_h+(1)/(2) V_h$

$V_(2f)=2V_h$

$V_(2f)=2\sqrt{2gh$

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity $V_f$ = 0

$V^2=V_(2f)^2-2gy$

$0=(2√(2gh))^2-2gy$

$y=4H$

Thus

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

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Answer 2

Answer:

Answer:

a) the large sphere has 3 times the mass of the small sphere

b) y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

vf² = vo² - 2g y

As the spheres are released v₀ = 0

vf² = - 2g H

vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

pf = m₁ v₁f + m₂ v₂f

po = pf

m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

Ko = KF

½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

-m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

mt = m₁ + m₂

v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

(m₁-m₂) / mt vh = 2 m₂ / mt vh

(m₁-m₂) = 2m₂

m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

mt = 3m2 + m2

mt= 4m2

v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

v₂f = 3/2 vh +1/2 vh

v₂f = 2 vh

v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

V² = v₂f² - 2 g Y

0 = (2√2gh)² - 2gy

2gy = 4 (2gH)

y = 4H

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A long, thin superconducting wire carrying a 17 A current passes through the center of a thin, 3.0-cm-diameter ring. A uniform electric field of increasing strength also passes through the ring, parallel to the wire. The magnetic field through the ring is zero.a. At what rate is the electric field strength increasing?
b. is the electric field in the direction of the current or opposite to the current?

Answers

Answer:

$(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The direction of the electric field is opposite that of the current

Explanation:

From the question we are told that

The current is $I = 17\ A$

The diameter of the ring is $d = 3.0 \ cm = 0.03 \ m$

Generally the radius is mathematically represented as

$r = (d)/(2)$

$r = (0.03)/(2)$

$r = 0.015 \ m$

The cross-sectional area is mathematically represented as

$A = \pi r^2$

=> $A = 3.142 * (0.015^2)$

=> $A = 7.07 *10^(-4 ) \ m^ 2$

Generally according to ampere -Maxwell equation we have that

$\oint \= B \cdot \= ds = \mu_o I + \epsilon_o \mu _o( d \phi )/(dt )$

Now given that $\= B = 0$ it implies that

$\oint \= B \cdot \= ds = 0$

$\mu_o I + \epsilon_o \mu _o( d \phi )/(dt ) = 0$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^(-12 ) \ m^(-3) \cdot kg^(-1)\cdot s^4 \cdot A^2$

$\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^(-7) N/A^2$

$\phi$ is magnetic flux which is mathematically represented as

$\phi = E * A$

Where E is the electric field strength

$\mu_o I + \epsilon_o \mu _o ( d [EA] )/(dt ) = 0$

=> $(dE)/(dt) =- (I)/(\epsilon_o * A )$

=> $(dE)/(dt) =- (17)/(8.85*10^(-12) * 7.07*10^(-4) )$

=> $(dE)/(dt) =- 2.72 *10^(15) \ N/C \cdot s$

The negative sign shows that the direction of the electric field is opposite that of the current

Which statementsabout a neutral atom are correct? Check all that apply.1. A neutral atom is composed of bothpositively and negatively charged particles.
2. Positively charged protons are located in the tiny, massivenucleus.
3. The positively chargedparticles in the nucleus are positrons.
4. The negatively chargedelectrons are spread out in a "cloud" around thenucleus.
5. The electrons areattracted to the positively charged nucleus.
6. The radius of the electroncloud is twice as large as the radius of the nucleus.

Answers

Answer:

1, 2, 4, 5 are correct

Explanation:

1) This is true because In a neutral atom, the number of positive charges (protons) is equal to the number of negative charges (electrons).

2) This is true because the mass of the atom which is made up of the protons and neutrons, is located in the tiny nucleus.

3) This is not true because the positively charged particles in the nucleus are called protons.

4) This is true because electrons move around the nucleus in diffuse areas known as orbitals.

5) This is true because opposite charges attract each other. And electron is a negative charge.

6) This is not true because the radius of the electron cloud is normally 10,000 times larger than the radius of the nucleus.

A 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V and negligible internal resistance. At the instant when the resistor is dissipating electrical energy at a rate of 300 W, how much energy has been stored in the capacitor?

Answers

Answer:

The energy stored is $E_s = 0.0064 \ J$

Explanation:

From the question we are told that

The capacitance is $C = 8 \ \mu F = 8*10^(-6) \ F$

The resistance is R = 3.00-Ω

The emf is $E_t = 70.0 V$

The power is P = 300 W

Generally the total emf is mathematically represented as

$E_t = E_c + E_r$

Here $E_c$ is the emf across that capacitor which is mathematically represented as

$E_c = (q)/(C)$

and $E_r$ is the emf across the resistor which is mathematically represented as

$E_r = √(P R)$

$E_t = √(PR) + (q)/(C)$

=> $q = C[E_t - √(PR) ]$

Generally the energy stored in a capacitor is mathematically represented as

$E_s = (q^2)/(2C)$

=> $E_s = ([C [ E_t - √(PR) ]]^2)/(2C)$

=> $E_s = ([8.0*10^(-6) [ 70 - √(300 * 3))/(2 *(8.0*10^(-6)))$

=> $E_s = 0.0064 \ J$

Final answer:

The energy stored in the capacitor is 0 J.

Explanation:

When a 8.00-μF capacitor that is initially uncharged is connected in series with a 3.00-Ω resistor and an emf source with E = 70.0 V

At the instant when the resistor is dissipating electrical energy at a rate of 300 W, we can calculate the current flowing through the circuit using Ohm's law: I = V/R = 70.0 V / 3.00 Ω = 23.33 A.

The energy stored in a capacitor can be calculated using the formula: E = 1/2 * C * V^2, where C is the capacitance and V is the voltage across the capacitor.

Since the capacitor is initially uncharged, the voltage across it is also zero. So the energy stored in the capacitor is 0.5 * 8.00 x 10^-6 F * (0 V)^2 = 0 J.

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While testing at 30 feet below the surface in Lake Minnetonka, with the sub stopped and in equilibrium, one of the students aboard the sub drops a hammer that goes through the hull of the submarine, and sticks out of the submarine handle first. When this happens, a seal forms immediately around the handle, so that no water enters the sub. What is the new equilibrium position for the sub?

Answers

Answer:

Explanation:

The equilibrium position of the sub is at the surface of the lake

Two point charges have a total electric potential energy of -24 J, and are separated by 29 cm.If the total charge of the two charges is 45 μC, what is the charge, in μC, on the positive one?
What is the charge, in μC, on the negative one?

Answers

Answer:

The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Explanation:

Electric potential energy between two point charges is derived from concept of Work, Work-Energy Theorem and Coulomb's Law and described by the following formula:

$U_(e) = (k\cdot q_(A)\cdot q_(B))/(r)$ (1)

Where:

$U_(e)$ - Electric potential energy, measured in joules.

$q_(A)$ , $q_(B)$ - Electric charges, measured in coulombs.

$r$ - Distance between charges, measured in meters.

$k$ - Coulomb's constant, measured in kilogram-cubic meters per square second-square coulomb.

If we know that $U_(e) = -24\,J$ , $q_(A) = 45* 10^(-6)\,C+ q_(B)$ , $k = 9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2))$ and $r = 0.29\,m$ , then the electric charge is:

$-24\,J = -(\left(9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2)) \right)\cdot (45* 10^(-6)\,C+q_(B))\cdot q_(B))/(0.29\,m)$

$-6.96 = -405000\cdot q_(B)-9* 10^(9)\cdot q_(B)^(2)$

$9* 10^(9)\cdot q_(B)^(2)+405000\cdot q_(B) -6.96 = 0$ (2)

Roots of the polynomial are found by Quadratic Formula:

$q_(B,1) = 1.327* 10^(-5)\,C$ , $q_(B,2) \approx -5.827* 10^(-5)\,C$

Only the first roots offer a solution that is physically reasonable. The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

What are some of the benefits of learned optimism that have been found inresearch?
O
A. Fewer health problems
O
O
B. All of these
C. Making more money
O
D. A lower divorce rate

Answers

The benefits of learned optimism that have been found in research are Fewer health problems, Making more money, and a lower divorce rate. The correct option is B.

Learned optimismhas been associated with numerous benefits in research, including fewer health problems, making more money, and a lower divorce rate. Optimistic people tend to have better physical and mental health, which leads to fewer health problems. Additionally, optimistic people tend to be more successful in their careers and finances, which can lead to higher income and better financial stability. Finally, optimistic people tend to have better relationships, including lower divorce rates, as they are better able to handle conflicts and maintain positive attitudes toward their partners.

In summary, learned optimism has a range of benefitsfor individuals, including better physical and mental health, greater success in work and education, better relationships with others, and improved resilience. These benefits make learned optimism an important skill for individuals to develop in order to lead happier, healthier, and more successful lives.

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