9) A balloon is charged with 3.4 μC (microcoulombs) of charge. A second balloon 23 cm away is charged with -5.1 μC of charge. The force of attraction / repulsion between the two charges will be: ______________________ 10) If one of the balloons has a mass of 0.084 kg, with what acceleration does it move toward or away from the other balloon? (calculate both magnitude AND direction) ________________________________________

Answer

given,

distance of fall = h

initial speed = 0 m/s

it travels 0.540 h in the last 1.00 s

Speed of the fall = [tex]\dfrac{Distance}{time}[/tex]

Speed = [tex]\dfrac{0.540\ h-0}{1}[/tex]

S = 0.540h m/s

time of the fall

using equation of motion

v = u + g t

0.540 h = 0 + 9.8 t

t = 0.055h s

The time of fall is equal to 0.055h s

height of fall = ?

again using equation of motion

v² = u² + 2 g s

(0.540h)² = 0 + 2 x 9.8 x s

s = 0.0148 h²

Hence, the time of fall of the object = 0.055h s.

            the distance of fall of object = 0.0148 h²

The object has been falling for a total time of 3.13 seconds and the total height from which it fell is 47.9 meters, given that it covered 0.540 of its total height in the last 1 second of free fall.

Let's tackle this physics problem step by step to find both the time the object has been falling and the total height from which it fell. We'll use the known acceleration due to gravity (g = 9.81 m/s²) and apply kinematic equations.

Given that the object travels 0.540h in the last 1.00 s, we can set up two equations using the kinematic formula h = ½gt², where h is the height, g is the acceleration due to gravity, and t is the time in seconds:

We solve the second equation for t, which will then be used to find the total height using the first equation.

Solving the second equation: 0.540h = ½g(t²) - ½g((t-1)²)

Expand and simplify the equation: 0.540h = ½g(t²) - ½g(t² - 2t + 1)

0.540 = t² - ½g(t² - 2t + 1)

Now, assuming g is approximately 9.8 m/s², we can plug in values and solve for t:

0.540 = t² - 0.5(9.81)(t² - 2t + 1)

After solving for t, we find:

This calculation assumes that air resistance is negligible and the acceleration due to gravity is constant.

Answer:

-3.63 degree Celsius

Explanation:

We are given that

Boiling point of solution=[tex]T_b=101^{\circ}[/tex] C

Boiling point water=100 degree Celsius

[tex]K_f=1.86K/m[/tex]

[tex]K_b=0.512 K/m[/tex]

[tex]\Delta T_b=T-T_0[/tex]

Where [tex]T[/tex]=Boiling point of solution

[tex]T_0=[/tex]Boiling point of pure solvent

[tex]\Delta T_b=101-100=1^{\circ}[/tex]C

[tex]\Delta T_b=k_bm[/tex]

Using the formula

[tex]1=0.512\times m[/tex]

Molality,[tex]m=\frac{1}{0.512}[/tex] m

[tex]\Delta T_f=k_fm[/tex]

Using the formula

[tex]\Delta T_f=\frac{1}{0.512}\times 1.86[/tex]

[tex]\Delta T_f=3.63 C[/tex]

We know that

[tex]\Delta T_f=T_0-T_1[/tex]

Where [tex]T_0[/tex] =Freezing point of solvent

[tex]T_1=[/tex] Freezing point of solution

Using the formula

[tex]3.63=0-T_1[/tex]

Freezing point of water=0 degree Celsius

[tex]T_1=0-3.63=-3.63 C[/tex]

Hence, the freezing point of solution=-3.63 degree Celsius

Final answer:

The freezing point of an aqueous solution can be calculated using the freezing point depression equation ΔTf = Kf * m. In this case, the solution boils at 101°C, indicating the boiling point elevation constant (Kb) is provided. Assuming complete dissociation, we can calculate the freezing point depression to be -0.512°C.

Explanation:

The freezing point depression can be calculated by using the equation:

ΔTf = Kf * m

Where ΔTf is the change in freezing point, Kf is the cryoscopic constant, and m is the molality of the solution.

In this case, since we have the boiling point elevation constant (Kb), we need to use the equation:

ΔTf = Kb * m

Given that Kb is 0.512 K/m and assuming complete dissociation, a 1.0 m aqueous solution of a solute will contain 1.0 mol of particles per kilogram of water. Therefore, the freezing point depression (ΔTf) will be:

ΔTf = 0.512 K/m * 1.0 m = 0.512 K

Since the freezing point of pure water is 0°C, the freezing point of the aqueous solution will be:

0°C - 0.512 K = -0.512°C

To find out how far the firefighting crew should position their cannon from the building, we use the equations of projectile motion separately for the vertical and the horizontal components. Since there are two viable times that satisfy the conditions of motion, there are two potential positions of the cannon. This question combines the principles of physics, which are specifically related to projectile motion.

This question is about the physics of projectile motion, specifically under gravity. To find out how far from the building they should position their cannon, we have to deal with the vertical and horizontal components of the projectile motion separately.

First, we determine the time it takes for the water to reach the desired height. We get this by using the equation of motion: H = V*sin(theta)*t - 0.5*g*t^2, where H is the height above the ground level (10m), V is speed (25.0 m/s), sin(theta) is the vertical component of initial velocity, g is acceleration due to gravity (approximated as 10 m/s^2 for ease of calculation), and t is time. We can rearrange this equation to find t.

Next, we use the time obtained and the horizontal component of initial velocity (V*cos(theta)) to find the horizontal distance made by using the equation: s = V*cos(theta)*t.

The hint here is that there are two potential time values, a smaller one and a larger one, leading to two potential distances from the building that the cannon could be positioned. Both times satisfy the conditions of the motion.

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The firefighting crew should place their cannon either 8.0 meters or 56.1 meters from the building to reach a blaze 10.0 meters above ground.

To determine how far the firefighting crew should position their cannon, we need to analyze the projectile motion of the water. We'll break this down into x-direction (horizontal) and y-direction (vertical) components.

Step-by-Step Solution

1. Break down initial velocity into components

Initial speed, v₀ = 25.0 m/s

Angle, θ = 53.0°

v₀x = v₀ x cos(θ) = 25.0 x cos(53.0°) ≈ 15.0 m/s

v₀y = v₀ x sin(θ) = 25.0 x sin(53.0°) ≈ 20.0 m/s

2. Write the equations of motion

Vertical (y-direction): y = v₀y x t - 0.5 x g x t²

Given y = 10.0 m, v₀y = 20.0 m/s, g = 9.8 m/s²

Using the quadratic formula, we find the time (t) it takes for the water to reach 10.0 m:

10.0 = 20.0 x t - 0.5 x 9.8 x t²

4.9t² - 20.0t + 10.0 = 0

Solving for t, we get t ≈ 0.54 s or t ≈ 3.74 s

Both times are valid because water can take two paths: one on the way up (0.54 s) and one on the way down (3.74 s).

3. Calculate horizontal distance

Distance (x) = v₀x x t

For t ≈ 0.54 s: x = 15.0 x 0.54 ≈ 8.0 m

For t ≈ 3.74 s: x = 15.0 x 3.74 ≈ 56.1 m

Therefore, the cannon can be placed either 8.0 meters or 56.1 meters from the building.

To solve this problem we will apply the concepts of the potential difference such as the product between the electric field and the distance, then we will use two definitions of capacitance, the first depending on the Area and the second depending on the load to find the Area. Finally we will look for capacitance with the values already obtained in the first sections of this problem

PART A) Potential Difference is

[tex]V = Ed[/tex]

Here,

E = Electric Field

d = Distance

Replacing,

[tex]V = (5*10^6)(3.0*10^{-3})[/tex]

[tex]V = 15000V= 15kV[/tex]

PART B) Capacitance of the capacitor is

[tex]C = \frac{\epsilon_0 A}{d}[/tex]

Here,

A = Area

[tex]\epsilon_0[/tex] = Permittivity Vacuum

d = Distance

Rearranging to find the Area we have,

[tex]A = \frac{Cd}{\epsilon_0}[/tex]

We know at the same time that Capacitance is the charge per Voltage, then

[tex]C = \frac{Q}{V}[/tex]

Replacing at this equation we have that

[tex]A = \frac{Qd}{V\epsilon_0}[/tex]

[tex]A = \frac{(79*10^{-9})(3*10^{-3})}{(15000)(8.853*10^{-12})}[/tex]

[tex]A = 1.78mm^2[/tex]

PART C)

Capacitance is given by,

[tex]C = \frac{Q}{V}[/tex]

[tex]C =\frac{79*10^{-9}}{15000}[/tex]

[tex]C =5.26pF[/tex]

Answer:

42 minutes

Explanation:

First, let us find out the time required by the roommate, who is driving the truck, to reach the city.

We know that,

[tex]time=\frac{distance}{speed} =\frac{360}{50} hrs=7.2hrs[/tex]

Now, you are planning to rest a bit after traveling for an hour. So,

distance covered in that 1 hour = [tex]60mi/h\times1h=60 miles[/tex]

time required by you to cover the total distance = [tex]\frac{360mi}{60mi/h} =6hours[/tex]

If you wish reach at the destination half an hour (0.5 h) before your roommate, you can expend a total of [tex]7.2-0.5=6.7hours[/tex] throughout your journey.

Hence, you can rest for  [tex]6.7-6=0.7hours=(0.7\times60)minutes=42minutes[/tex]

After driving 60 miles in the first hour and intending to arrive half an hour before your roommate who travels at 50 mph, you can afford a break time of 30 minutes at the rest stop.

The subject of your question is

relative speed

and

time management

, which falls under Mathematics. Your overall travelling speed is 60 mph. After driving for an hour, you decide to take a break at a rest stop. In that one hour, you've driven 60 miles, so there are 300 miles left to the destination.  Your roommate, who doesn't stop for a break, continues to drive at 50 mph. You want to arrive half an hour before she does, which means you essentially have her travel time less 30 minutes to reach the destination. Because her speed in relation to the remaining distance is constant, her remaining travel time is 300 miles divided by 50 mph which equals to 6 hours. So, you actually have 6 - 0.5 = 5.5 hours to travel the remaining 300 miles including your rest stop. Considering your speed of 60 mph, it will take you 300/60 = 5 hours to reach the destination. Therefore, the length of the break you can take while still beating your roommate to the destination is 5.5 hours minus 5 hours, which is

30 minutes

.

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Answer:

T = 98.1 N

Explanation:

Given:

- mass of bar m = 10 kg

- Length of the bar L = 2 m

- Angle between Cable and wall Q = 30 degres

Find:

- Find the tension in the cable.

Solution:

- Take moments about intersection of bar and wall, to be zero (static equilibrium)

                                    (M)_a = 0

                                    T*sin( 30 )*L - m*g*L/2 = 0

                                    T*sin(30) - m*g / 2 = 0

                                    T = m*g / 2*sin(30)

                                    T = 10*9.81 / 2*sin(30)

                                    T = 98.1 N

To solve this problem we will rely on the theorems announced by Newton and Coulomb about the Gravitational Force and the Electrostatic Force respectively.

In the case of the Force of gravity we have to,

[tex]F_g = G\frac{m_pm_e}{d^2}[/tex]

Here,

G = Gravitational Universal Constant

[tex]m_p[/tex] = Mass of Proton

[tex]m_e[/tex] = Mass of Electron

d  = Distance between them.

[tex]F_g = (6.673*10^{-11} kg^{-1} \cdot m^3 \cdot s^{-2}) (\frac{(1.672*10^{-27}kg)(9.109*10^{-31})}{(52.9pm)^2})[/tex]

[tex]F_g = 3.631*10^{-47}N[/tex]

In the case of the Electric Force we have,

[tex]F_e = k\frac{q_pq_e}{d^2}[/tex]

k = Coulomb's constant

[tex]q_p[/tex] = Charge of proton

[tex]q_e[/tex] = Charge of electron

d = Distance between them

[tex]F_e = (9*10^9N\cdot m^2 \cdot C^{-2})(\frac{(1.602*10^{-19}C)(1.602*10^{-19}C)}{(52.9pm)^2})[/tex]

[tex]F_e = 82.446*10^{-9}N[/tex]

Therefore

[tex]\frac{F_e}{F_g} = 2.270*10^{39}[/tex]

We can here prove that the statement is True

It has been proved in below calculation that the gravitational force exerted by a proton on an electron is [tex]2\times10^{39}[/tex]times weaker than the electric force that the proton exerts on an electron

Gravitational force is the universal force of attraction acting between two bodies.

It is given by,

[tex]Fg=G\times\dfrac{m_1 \times m_2}{x^2}[/tex]

Here [tex]G[/tex] is gravitational constant [tex]m[/tex] is the mass of the bodies and [tex]x[/tex] is the distance between bodies.

Electric force is the force of attraction or repulsion acting between two charged bodies,

[tex]F_E=k\times\dfrac{q_1 \times q_2}{x^2}[/tex]

Here, [tex]k[/tex] is Coulomb's constant [tex]q[/tex] is the charge and [tex]x[/tex] is the distance between bodies.

The gravitational force exerted by a proton on an electron is,

[tex]Fg=6.673\times{10^{-11}}\times\dfrac{1.673\times10^{-27}\times9.1094\times10^{-31}}{x^2}[/tex]

The electric force exerted by a proton on an electron is,

[tex]F_E=9\times{10^{9}}\times\dfrac{1.602\times10^{-19}\times1.602\times10^{-19}}{x^2}[/tex]

Compare both,

[tex]\dfrac{F_g}{F_e} =\dfrac{6.673\times{10^{-11}}\times\dfrac{1.673\times10^{-27}\times9.1094\times10^{-31}}{x^2}}{9\times{10^{9}}\times\dfrac{1.602\times10^{-19}\times1.602\times10^{-19}}{x^2}}\\\dfrac{F_g}{F_e} =\dfrac{1}{2\times10^{39}}[/tex]

Thus, It has been proved in below calculation that the gravitational force exerted by a proton on an electron is [tex]2\times10^{39}[/tex]times weaker than the electric force that the proton exerts on an electron.

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Answer with Explanation:

We are given that

r=3 m

Angular frequency=[tex]\omega[/tex]=560rev/min

A.1 revolution=[tex]2\pi[/tex] radian

560 revolutions=[tex]560\times 2\pi[/tex] rad

Angular frequency=[tex]2\times 3.14\times \frac{560}{60}[/tex]rad/s

1 min=60 s

[tex]\pi=3.14[/tex]

Angular frequency=[tex]\omega=58.6rad/s[/tex]

Linear speed=[tex]\omega r[/tex]

Using the formula

Linear speed=[tex]58.6\times 3=175.8m/s[/tex]

Hence, the linear speed of the blade tip=175.8m/s

B.Radial acceleration=[tex]a_{rad}=\frac{v^2}{r}[/tex]

By using the formula

Radial acceleration=[tex]a_{rad}=\frac{(175.8)^2}{3}= 10.301\times 10^3m/s^2[/tex]

Radial acceleration=[tex]\frac{10.301}{9.8}g\times 10^3=1.05\times 10^3 g[/tex]

Where [tex]g=9.8m/s^2[/tex]

Hence, the radial acceleration[tex]=1.05\times 10^3 g[/tex]

Answer:

The force on the electron will become zero.

Explanation:

As electron has negative charge and proton has positive charge. The magnitude of charge on both particles is equal. Therefore, there will be a force of attraction between electron and proton. When another electron is brought near the proton the net charge in that area will become equal to zero. Therefore, first electron will not experience any force.

Final answer:

The original force between the first electron and proton remains unchanged when a second electron is placed next to the proton, but the first electron will experience an additional repulsive force from the second electron. This alters the net force acting on the first electron but does not directly change the force between it and the proton.

Explanation:

The question inquires about the effect on the force on an electron when a second electron is placed next to a proton, with all particles separated by a distance of 1 meter.

According to Coulomb's Law, which governs the electrostatic force between two charged particles, the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

When a second electron is placed next to the proton, this setup introduces additional forces: a repulsive force between the two electrons and an attractive force between the new electron and the proton.

However, the original question seems to focus on how the force on the first electron changes with this new arrangement. The presence of a second electron does not alter the magnitude of the force between the first electron and the proton directly since the forces here are evaluated pairwise, and the distance between them remains unchanged.

What changes, though, is the overall electrical environment, introducing a new set of forces that need to be considered for the complete system. The addition of the second electron introduces a repulsive force on the first electron, which is separate from but concurrent with the attractive force from the proton.

It is important to consider the net force acting on any charge in such scenarios, which would involve adding vectorially the attractive force from the proton and the repulsive force from the second electron.

Thus, while the force due to the proton-electron pair remains constant, the first electron experiences an additional repulsive force due to the second electron, affecting the net force on it but not the force between it and the proton directly.

An atom in normal conditions refers when electrons are in the fundamental state. When you leave the atom, an electron absorbs energy from an external source and moves to a higher energy state.

Energy states or energy levels are called orbitals. The difference between the energy states in an atom is responsible for the emission of photons when an electron transition occurs between these two energy states.

Because the energy levels are discrete, the emitted photons also possess different energies.

Answer:

Explanation:

Given

Mass of big box is M and small box is m

Tension T will cause the boxes to accelerate

[tex]T=(M+m)a[/tex]

where a=acceleration of the boxes

Now smaller box will slip over large box if the acceleration force will exceed the static friction

i.e. for limiting value

[tex]\mu _smg=ma[/tex]

[tex]a=\mu _s\cdot g[/tex]

thus maximum tension

[tex]T=\mu _s(M+m)g[/tex]

Answer:

(a) 11.66 square meters per liter

(b) 11657.8 per meters

(c) 0.00211 gal per square feet

Explanation:

(a) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 1gal/3.7854L = 11.66m^2/L

(b) 475ft^2/gal = 475ft^2/gal × (1m/3.2808ft)^2 × 264.17gal/1m^3 = 11657.8/m

(c) Inverse of 475ft^2/gal = 1/475ft^3/gal = 0.00211gal/ft^3

To solve this problem we will apply the concepts related to gravitational potential energy.

This can be defined as the product between mass, gravity and body height.

Mathematically it can be expressed as

[tex]\Delta P = mgh[/tex]

[tex]\Delta P = (8.7)(9.8)(3)[/tex]

[tex]\Delta P = 255.78J[/tex]

Therefore the change in the internal energy of the system is 255.78

Answer:

The maximum volume flow rate is 0.03745m^3/s

Explanation:

Power input (Pi) = 9-hp = 9×746W = 6,714W

Elevation (h) = 15m

Efficiency (E) = 82% = 0.82

Density (D) of water = 1000kg/m^3

E = Po/Pi

Po (power output) = E×Pi = 0.82×6,714W = 5505.48W

Po = mgh/t

m/t (mass flow rate) = Po/gh = 5505.48/9.8×15 = 37.45kg/s

Volume flow rate = mass flow rate ÷ density = 37.45kg/s ÷ 1000kg/m^3 = 0.03745m^3/s

Answer:

a)[tex]q=14.5\ W/m^2[/tex]

b)Q= 58 W  

Explanation:

Given that

Thermal conductivity ,K = 0.029 W/m.k

The temperature difference ,ΔT= 10°C

The thickness ,L = 20 mm

We know that

[tex]Q=\dfrac{KA}{L}\times \Delta T[/tex]

Now by putting the values

[tex]Q=\dfrac{0.029\times 4}{0.02}\times 10\ W[/tex]

Q= 58 W

The heat flux through the sheet is given as

[tex]q=\dfrac{Q}{A}\ W/m^2[/tex]

[tex]q=\dfrac{58}{2\times 2}\ W/m^2[/tex]

[tex]q=14.5\ W/m^2[/tex]

a)[tex]q=14.5\ W/m^2[/tex]

b)Q= 58 W

Explanation:

The wavelength of green light is about 500 nanometers, or two thousandths of a millimeter. The typical wavelength of a microwave oven is about 12 centimeters, which is larger than a baseball.

The statement is falls, because the the wavelength of green light is about 500 nm or 500 × 10⁻⁹ m. But the size of an atom is  about 1.2  × 10⁻¹⁰ m. Hence atomic size not equals the wavelength of green light.

The wavelength of an electromagnetic wave is the distance between two consecutive crests or troughs.  The upward peak in the wave is called crests and the downward peaks are called troughs.

The wavelength of high energy waves will be shorter. Visible region is in between IR and UV rays in the electromagnetic spectrum. Green light is in the exact middle region of the VIBGYOR. Thus, it is having a wavelength of 500 -520 nm or  500 × 10⁻⁹ m.

The size of an atom is estimated in the range of 1.2  × 10⁻¹⁰ m and it varies from element to element. However the atomic size is not comparable with the wavelength of green light.

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The electric flux through one of the faces of the cube with the charge at its geometric center can be found using Gauss's Law. The electric flux  enclosed by the surface divided by the permittivity of the medium is 1.424 N·m2/C.

We can use Gauss's Law to determine the electric flux through one of the faces of the cube with the charge at its geometric center. Gauss's Law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of the medium.

The charge enclosed by one face of the cube is 12.6 µC. The permittivity of a vacuum is 8.85419 × 10-12 C2/N·m2.

Therefore, the electric flux through one face of the cube is (12.6 µC) / (8.85419 × 10-12 C2/N·m2) = 1.424 N·m2/C.

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The batteries do not make the charges move as much as the hand generator. This is because batteries store energy chemically, and depending on their properties, they may not provide enough energy to power all the bulbs in the circuit. In contrast, hand generators can generate a higher voltage due to a greater induced EMF from rapid movement in a magnetic field.

The best explanation for this phenomenon is that 'the batteries do not make the charges move as much' as compared to the hand generator. Hand generators work on the principle of electromagnetic induction where the rapid movement of a coil in a magnetic field can induce a substantial electromotive force (emf). This high emf allows the generator to power more light bulbs in a circuit. This is because the faster the generator is spun, the greater the induced emf, which in turn produces a higher voltage.

On the other hand, the batteries store energy as a chemical reaction waiting to happen, not as electric potential. This reaction only runs when a load is attached to both terminals of the battery. Hence, depending on the energy storage and discharge properties of the battery used, they may not be able to provide enough energy to sufficiently power all the bulbs in a circuit, especially if they are made to power more than one bulb simultaneously. Hence, in the case of the two batteries, they are not able to cause the charges to move as much as the hand generator, and this results in a lower voltage and consequently dimmer lights.

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The distance from the starting point is approximately 26.3 m and the compass direction is 25.3° west of north.

To find the distance between the starting point and the final position, we can use the Pythagorean theorem. We can calculate the distances in the north and west directions using the given lengths and angles. Using the given values, the distance from the starting point is approximately 26.3 m. The compass direction of the line connecting the starting point to the final position is 25.3° west of north.

Final answer:

To find the distance between the starting point and final position, use the Pythagorean theorem. The total distance is approximately 31.3 m. To determine the compass direction, use trigonometry. The compass direction is approximately 36.9° north of west.

Explanation:

To find the distance between your starting point and final position, you can use the Pythagorean theorem. First, calculate the horizontal distance by adding the horizontal components of the two legs of the walk: A = 18.0 m west and B = 0 m south. The horizontal distance is 18.0 m. Then, calculate the vertical distance by summing up the vertical components of the two legs of the walk: A = 0 m north and B = 25.0 m north. The vertical distance is 25.0 m. Apply the Pythagorean theorem to find the total distance by taking the square root of the sum of the squares of the horizontal and vertical distances. The total distance is approximately 31.3 m.

To determine the compass direction of a line connecting your starting point to your final position, you can use trigonometry. First, calculate the angle θ between the horizontal distance (18.0 m) and the hypotenuse (31.3 m). You can use the inverse tangent function: θ = arctan(vertical distance/horizontal distance). Calculate θ to be approximately 53.1°. Since you walked west and then north, the compass direction is 90° less than θ. Therefore, the compass direction is approximately 36.9° north of west.

To solve this problem we will apply the concepts of equilibrium and Newton's second law.

According to the description given, it is under constant ascending acceleration, and the balance of the forces corresponding to the tension of the rope and the weight of the elevator must be equal to said acceleration. So

[tex]\sum F = ma[/tex]

[tex]T-mg = ma[/tex]

Here,

T = Tension

m = Mass

g = Gravitational Acceleration

a = Acceleration (upward)

Rearranging to find T,

[tex]T = m(g+a)[/tex]

[tex]T = (975)(9.8+0.754)[/tex]

[tex]T= 10290.15N[/tex]

Therefore the tension force in the cable is 10290.15N

Answer: time(t) = 1.72s

Explanation: [tex]Angular acceleration =\frac{angular velocity}{time taken}[/tex]

Let angular acceleration = α = [tex]2.69rad/s^{2}[/tex]

ω = angular velocity = 44rev/ min

the angular velocity is in rev/ min but we need to have it in rad/s , thus we do so below

recall that 1 rev = 2π  and 1 min = 60s, [tex]\frac{44.2 * 2\pi }{60} \\\\ \frac{44.2 * 2*3.142}{60} \\\\= 4.63rads^{-1}[/tex]

hence 44.2rev/min = 44.2 * 2π/ 60 = 44.2 * 2 *3.142/ 60

thus angular velocity = 4.63rad/s

time taken = angular velocity/ angular acceleration

time taken = 4.63/2.69

time taken = 1.72s

Answer:

a)The keys were thrown with an initial velocity of 10.0 m/s.

b)The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).

Explanation:

Hi there!

The equations for the height and velocity of the keys are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h =height of the keys after a time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² considering the upward direction as positive)

v = velocity of the keys at time t.

a) We know that at t = 1.50 s, h = 4.00 m and let´s consider that the origin of the frame of reference is located at the point where the keys are thrown so that h0 = 0. Then, using the equation of height, we can obtain the initial velocity.

h = h0 + v0 · t + 1/2 · g · t²

4.00 m = v0 · 1.50 s - 1/2 · 9.81 m/s² · (1.50 s)²

4.00 m + 1/2 · 9.81 m/s² · (1.50 s)² = v0 · 1.50 s

v0 = ( 4.00 m + 1/2 · 9.81 m/s² · (1.50 s)²) / 1.50 s

v0 = 10.0 m/s

The keys were thrown with an initial velocity of 10.0 m/s.

b) Now, using the equation of velocity we can calculate the velocity at t = 1.50 s:

v = v0 + g · t

v = 10.0 m/s - 9.81 m/s² · 1.50 s

v = -4.72 m/s

The velocity of the keys just before they were caught is -4.72 m/s (the keys were caught on their way down).

To solve this problem we will apply the principles of energy conservation. The kinetic energy in the object must be maintained and transformed into the potential electrostatic energy. Therefore mathematically

[tex]KE = PE[/tex]

[tex]\frac{1}{2} mv^2 = \frac{kq_1q_2}{r}[/tex]

Here,

m = mass (At this case of the proton)

v = Velocity

k = Coulomb's constant

[tex]q_{1,2}[/tex] = Charge of each object

r= Distance between them

Rearranging to find the second charge we have that

[tex]q_2 = \frac{\frac{1}{2} mv^2 r}{kq_1}[/tex]

Replacing,

[tex]q_2 = \frac{\frac{1}{2}(1.67*10^{-27})(3*10^5)^2(7*10^{-2})}{(9*10^9)(1.6*10^{-19})}[/tex]

[tex]q_2 = 3.6531nC[/tex]

Therefore the charge on the sphere is 3.6531nC

The correct equation to use to find the distance 's' when the iron block and magnet come to a rest is ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation represents the conversion of potential energy into work done against friction.

((mB)^2 + mM)gH = μ(mB + mM)gs.

This question represents a problem of mechanics in Physics, specifically involving both potential energy and work-energy theorem. The relevant equation to use in finding the distance 's' in this scenario is: ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation derives from setting the initial potential energy of the system equal to the final kinetic energy when it rests, including the energy dissipated due to friction.

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The correct equation to find 's' when an iron block slides down a hill and collides with a magnet before encountering a rough surface, is ((mB)^2 + mB + mM)gH = μ(mB + mM)gs. This equation is derived from equating the kinetic energy at the base of the hill with the work done by friction.

The correct equation to use to define s in this scenario would be: ((mB)2mB+mM)gH=μ(mB+mM)gs. This equation stems from equating the energy at the top of the hill (kinetic + potential energy) with the work done against friction (which eventually stops the two masses). Let's understand it in a detailed step-by-step manner.

Step 1: At the base of the hill, the kinetic energy of the block is equal to the potential energy at the top of the hill, which can be represented as (mB + mM)gH = 0.5(mB + mM)v^2. From this, we can turn the speeds into velocities, which in turn gives us v = sqrt(2gH).

Step 2: When the two masses encounter the rough surface, they will lose their kinetic energy due to friction. The total kinetic energy lost, which is equal to work done by friction, can be represented as 0.5(mB + mM)v^2 = μ(mB + mM)gs. Substituting the value of v from step 1 into this equation gives us the final formula: ((mB)^2 + mB + mM)gH = μ(mB + mM)gs.

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Answer: the force exerted by the book on the earth

The reaction force to the force of gravity acting on a book resting on a shelf is the force exerted by the shelf, also known as the normal force. This force counteracts the force of gravity and is equal to the weight of the book. Understanding this concept requires knowledge of Newton's Second and Third Laws of Motion.

The reaction force to the force of gravity acting on a book resting on a shelf is the Force exerted by the shelf on the book. The shelf applies a Normal force, perpendicular to its surface, that counteracts the force of gravity pulling down on the book. This is directly tied into Newton's Third Law, which states that for every action, there is an equal and opposite reaction.

The normal force exerted by the shelf is a type of contact force and is exactly equal to the weight of the book (which is the force of gravity acting on the book). If the normal force were weaker, the book would begin to sink into the shelf. If it were stronger, the book would start to lift off of the shelf. Frictional force also plays a role here, preventing the book from sliding off the shelf, but it is not the reaction force to gravity in this case. Newton's Laws of Motion, particularly the Second and Third Laws, are key to understanding these dynamics.

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Explanation:

Expression for energy balance is as follows.

        [tex]\Delta E_{system} = E_{in} - E_{out}[/tex]

or,          [tex]E_{in} = E_{out}[/tex]

Therefore,  

         [tex]m(h_{1} \frac{v^{2}_{1}}{2}) = m (h_{2} \frac{V^{2}_{2}}{2})[/tex]

          [tex]h_{1} + \frac{V^{2}_{1}}{2} = h_{2} + \frac{V^{2}_{2}}{2}[/tex]

Hence, expression for exit velocity will be as follows.

           [tex]V_{2} = [V^{2}_{1} + 2(h_{1} - h_{2})^{0.5}[/tex]

                      = [tex]V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]

As [tex]C_{p}[/tex] for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.

       [tex]V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}[/tex]                    

                  = [tex][(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}[/tex]

                 = 40.7 m/s

Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.

A star that is cooler than the sun would appear red (a), because the color of a star is associated with its temperature, according to Wien's Law.

The color of a star is closely related to its temperature. In the case of a star that's cooler than the sun, it would appear (a) red. This is due to a principle in astrophysics known as Wien's Law, which states that the peak wavelength (color) of the light emitted by an object (such as a star) shifts to longer, redder wavelengths as the object gets cooler. So, cool stars like red giants appear red, while hotter stars appear blue or white.

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Final answer:

Measuring 20 cycles of the pendulum motion to determine the period instead of just one cycle allows for a more accurate and representative measurement. By averaging out any errors or variations in the measurement process, the value for the period of the pendulum is more precise. Measuring multiple cycles also helps identify any potential changes in the period over time.

Explanation:

The reason we measure 20 cycles of the pendulum motion to determine the period, rather than just one cycle, is to obtain a more accurate and representative measurement of the period. By measuring multiple cycles, we can average out any errors or variations in the measurement process, resulting in a more precise value for the period of the pendulum.

For example, if we were to measure only one cycle of the pendulum, any small errors in the timing or counting of the cycles could significantly affect our measurement. However, by measuring 20 cycles and then dividing the total time by 20, we can minimize these errors and obtain a more reliable measurement of the period.

Furthermore, measuring multiple cycles allows us to observe any potential changes in the period over time. In certain situations, the period of a pendulum may vary slightly due to factors such as air resistance or changes in the length of the string. By measuring multiple cycles, we can identify and account for any such variations, providing a more comprehensive understanding of the pendulum's behavior.

Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

[tex]K_{rel}=(\gamma-1)mc^2[/tex]

[tex]\dfrac{K_{rel}}{mc^2}=\gamma-1[/tex]

Put the value into the formula

[tex]\gamma=\dfrac{105.7}{0.511}+1[/tex]

[tex]\gamma=208[/tex]

We need to calculate the electron’s velocity

Using formula of velocity

[tex]\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}[/tex]

[tex]\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}[/tex]

[tex]\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1[/tex]

[tex]v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2[/tex]

Put the value into the formula

[tex]v^2=\dfrac{1-(208)^2}{-208^2}\times c^2[/tex]

[tex]v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}[/tex]

[tex]v=0.9999 c\ m/s[/tex]

Hence, The electron’s velocity is 0.9999 c m/s.

Answer:

12.56 A.

Explanation:

The magnetic field of a conductor carrying current is give as

H = I/2πr ............................... Equation 1

Where H = Magnetic Field, I = current, r = distance, and π = pie

Making I the subject of the equation,

I = 2πrH............... Equation 2

Given: H = 1 T, r = 2 m.

Constant: π = 3.14

Substitute into equation 2

I = 2×3.14×2×1

I = 12.56 A.

Hence, the magnetic field = 12.56 A.