A cylinder is rotating about an axis that passes through the center of each circular end piece. The cylinder has a radius of 0.0700 m, an angular speed of 88.0 rad/s, and a moment of inertia of 0.850 kg · m2. A brake shoe presses against the surface of the cylinder and applies a tangential frictional force to it. The frictional force reduces the angular speed of the cylinder by a factor of two during a time of 4.40 s.(a) Find the magnitude of the angular deceleration of the cylinder. rad/s 2 (b) Find the magnitude of the force of friction applied by the brake shoe. N

The distance from the top of the wood block to the water can be calculated using the buoyancy force. For fresh water, the distance is 0.1 cm and for sea water, it is 0.09709 cm.

To determine the distance from the top of the wood block to the water, we need to consider the buoyancy force acting on the block. When an object is submerged in a fluid, it experiences an upward force called buoyant force equal to the weight of the fluid displaced by the object.

The density of the wood block is given as 700 kg/m^3. Since the block floats, the density of the wood is less than the density of water (1000 kg/m^3). Therefore, the volume of the wood block is the same as the volume of water it displaces.

To find the distance from the top of the block to the water, we first need to calculate the volume of the block. The volume formula for a cube is V = s^3, where s is the length of a side. In this case, the side length of the block is 10 cm, so the volume is 10 cm x 10 cm x 10 cm = 1000 cm^3.

Next, we can calculate the weight of the water displaced by the block. The weight of an object is given by the formula W = m x g, where m is the mass and g is the acceleration due to gravity. The mass of the displaced water can be found using the density formula m = d x V, where d is the density and V is the volume. In this case, the density of water is 1000 kg/m^3 and the volume of the block is 1000 cm^3. Converting the volume to m^3, we get 0.001 m^3.

Finally, we can calculate the distance from the top of the block to the water using the formula h = (W/Water density) x (100 cm/1 m). Plugging in the values, we get h = (0.001 m^3 x 1000 kg/m^3) / (1000 kg/m^3) x (100 cm/1 m) = 0.001 m x 100 cm = 0.1 cm.

Therefore, the distance from the top of the block to the water is 0.1 cm when the water is fresh. For sea water, the density is higher at around 1030 kg/m^3. Using the same calculations, we get h = (0.001 m^3 x 1000 kg/m^3) / (1030 kg/m^3) x (100 cm/1 m) = 0.001 m x 97.09 cm = 0.09709 cm.

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The student’s questions involve the principles of thermal radiation and heat conduction in Physics, relating to the determination of the equilibrium temperature in a radiative system and the electrical power needed to maintain a temperature. Calculations require additional data on materials and geometry to provide accurate answers.

The student's questions pertain to the topics of thermal radiation and heat conduction in Physics, which are within the realm of thermodynamics. Specifically, they are asking about (a) the equilibrium temperature of a conical surface and (b) the power required to maintain a certain temperature on a circular plate. These calculations involve understanding Stefan-Boltzmann law, conduction, and power calculations.

Unfortunately, without further information or input parameters for materials and geometries, providing a definitive answer for a system like the one described (circular plate and conical surface in an enclosure) isn't possible. To determine the electrical power required, one would need to know properties like thermal conductivity, surface area, and possibly the geometry of the objects in question to find out the heat transfer rates.

i)  The average resistive force acting on the car is approximately 2.4 × 10³ N. ii)  The kinetic energy of the car is 3,888 J. b) The average braking force exerted on the car is coming to rest is approximately 259.2 N. ii)  The average braking force exerted on the car is coming to rest is approximately 259.2 N. 2) c) The change in temperature of each rear brake is approximately 0.907 Kelvin.

i) Given information:

Mass of the car (m) = 960 kg

Speed of the car (v) = 9.0 m/s

The angle of the slope (θ) = 15°

Resistive force (Fr) = m × g × sin(θ)

Fr = 960 × 9.8 × sin(15°)

Fr = 2.4 × 10³ N

Therefore, the average resistive force acting on the car is approximately 2.4 × 10³ N.

(ii) To calculate the kinetic energy of the car, we can use the formula:

Kinetic energy (KE) = (1/2) × m × v²

KE = (1/2) × 960 × (9.0)²

KE = 3,888 J

Therefore, the kinetic energy of the car is 3,888 J.

(b) The average braking force exerted on the car is:

Average braking force = Change in kinetic energy/braking distance

Average braking force = KE / braking distance

Average braking force = 3,888 / 15

Average braking force = 259.2 N

Therefore, the average braking force exerted on the car in coming to rest is approximately 259.2 N.

(ii) The work done by a force can be calculated using the formula:

Work = Force × Distance × cos(θ)

3,888 = Force × 15 × cos(0°)

Since cos(0°) = 1, the equation simplifies to:

3,888 = Force × 15

Force = 3,888 ÷ 15

Force = 259.2 N

Therefore, the average braking force exerted on the car in coming to rest is approximately 259.2 N.

2) (c) To calculate the change in temperature of each rear brake, we can use the formula:

Change in heat energy = mass × specific heat capacity × change in temperature

Change in heat energy = Work

The work done by the braking force is equal to the product of the force and the braking distance (15 m),

Therefore, the change in heat energy for each brake is:

Change in heat energy = Work = Force × braking distance

Change in heat energy = 259.2 × 15 = 3,888 J

Now, we can use the formula to find the change in temperature:

Change in heat energy = mass × specific heat capacity × change in temperature

3,888  = 5.2 × 900 × change in temperature

change in temperature ≈ 0.907 K

Therefore, the change in temperature of each rear brake is approximately 0.907 Kelvin.

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The angular momentum of the projectile about the origin when the particle is at the following locations is zero

A projectile of mass m is launched with an initial velocity making an angle θ with the horizontal as shown below (attached). The projectile moves in the gravitational field of the Earth. The angular momentum of the projectile about the origin when the particle is at the following locations. (Use the following as necessary: vi, θ, m, and g for the acceleration due to gravity.)

[tex]L=r x p[/tex]

Angular momentum is zero because there is no r.  Since the angular momentum depends on then cross product of the position vector and momentum vector which is in the same direction as velocity, therefore the formula for the angular momentum can be written as [tex]mvr_{perpendicular}[/tex], where the cross product component of r vector along/parallel to velocity/momentum is 0.

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Planetary geologists suspect that the original surface of a certain celestial body has been covered by younger volcanic activity, based on the small number of impact craters. Evidence of past impact events is gradually erased by the constant renewal of Earth's crust through plate tectonics.

Based on the small number of impact craters, planetary geologists suspect that virtually all the original surface of a certain celestial body has been covered by younger volcanic activity.

This can be observed in the case of Earth, where the constant renewal of its crust through plate tectonics gradually erases evidence of past impact events.

Geologists have identified eroded remnants of impact craters on Earth, providing evidence for the influence of these impacts on the planet's evolution over time.

1. The weight of the suitcase is approximately 1448.32 N.

2. The apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.

To find the weight and mass of the suitcase, we'll use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration.

Given:

Force (F) = 105 N

Acceleration (a) = 0.710 m/[tex]s^2[/tex]

We can use the equation F = ma to solve for mass (m):

105 N = m * 0.710 m/s^2

Rearranging the equation, we have:

m = F / a

m = 105 N / 0.710 m/[tex]s^2[/tex]

m ≈ 147.89 kg

So, the mass of the suitcase is approximately 147.89 kg.

To find the weight of the suitcase, we can use the equation W = mg, where W is the weight and g is the acceleration due to gravity (approximately 9.8 m/[tex]s^2[/tex]).

W = 147.89 kg * 9.8 m/[tex]s^2[/tex]

W ≈ 1448.32 N

Therefore, the weight of the suitcase is approximately 1448.32 N.

Now let's calculate the apparent weight of the astronaut aboard the rocket.

Given:

Mass (m) = 86 kg

Acceleration (a) = 29.0 m/[tex]s^2[/tex]

Using the same formula W = mg, we can find the apparent weight (W) of the astronaut:

W = m * a

W = 86 kg * 29.0 m/[tex]s^2[/tex]

W ≈ 2504 N

Therefore, the apparent weight of the 86-kg astronaut aboard the rocket is approximately 2504 N.

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The amplitude is 2.30mm, frequency is 118 Hz, wavelength is 0.899 m, wave speed is 106 m/s, the wave is moving in the direction of decreasing x, tension in the rope is 28 N and the average power transmitted by the wave is approximately 0.38W.

The wave function of a travelling wave on a thin rope is given by y(x,t) = 2.30mm * cos[(6.98rad/m) * x + (742 rad/s) * t]. From this equation, we can derive the following information:

Amplitude - This is the maximum displacement of a point on the wave and is determined by the coefficient before the cosine in the equation. Therefore, the amplitude is 2.30mm.

Frequency - This can be obtained by dividing the wave speed by the wavelength. The coefficient in front of the 't' in the equation is equivalent to 2π times the frequency (since there are 2π rad in one wave cycle). Therefore, the frequency f = (742 rad/s)/(2π) = 118 Hz.

Wavelength - This is represented by the reciprocal of the coefficient in front of 'x' in the equation. Therefore, the wavelength λ = 2π/(6.98 rad/m) = 0.899 m.

Wave Speed - This can be found by multiplying frequency and wavelength together. Hence, wave speed v = f * λ = 118 Hz * 0.899 m = 106 m/s.

Direction the wave is travelling - The symbol '+' in front of 't' in the equation tells us that the wave is moving in the direction of decreasing x.

Tension in the rope - This can be found using the wave speed that we have already calculated and the rope's mass per unit length µ (mass/length = 0.00338 kg / 1.35 m = 0.00250 kg/m). The tension T can be calculated using the formula v = sqrt(T/µ), which gives T = µ * v² = 0.00250 kg/m * (106 m/s)² = 28 N.

Average power transmitted by the wave - The power P carried by wave is given by the formula P = 0.5 * µ * v * A² * ω², where A is the amplitude, ω is the angular frequency (which is the coefficient in front of 't' in the equation). Substituting the values we find: P = 0.5 * 0.00250 kg/m * 106 m/s * (2.30 mm)² * (742 rad/s)² ≈ 0.38 W.

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The width of the slit can be determined using the formula θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. By rearranging the formula and substituting the given values, we find that the width of the slit is approximately 244 nm.

The width of the slit can be determined using the formula for the angle of the first minimum in a single-slit diffraction pattern, which is given by θ = λ / (2d), where θ is the angle, λ is the wavelength, and d is the width of the slit. In this case, the angle is 45.0 degrees and the wavelength is 470 nm. Converting the angle to radians and rearranging the formula, we can solve for the width of the slit:

d = λ / (2 * tan(θ))

Substituting the given values, we have:

d = (470 nm) / (2 * tan(45.0 degrees))

Calculating this, we find that the width of the slit is approximately 244 nm.

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Stars do not crash into each other because the gravitational force acts as a centripetal force, maintaining circular orbits and keeping them apart. The masses of the stars can be calculated using Kepler's third law for circular orbits.

The scenario provided describes a system in which two equal-mass stars maintain a constant distance apart and revolve around their common center of mass due to the gravitational force between them. The system is in balance when the gravitational force provides the necessary centripetal force to keep the stars in their circular orbits without them crashing into each other.

Part (a)

The two stars don't crash into one another because the gravitational attraction is providing the centripetal force that keeps them in stable, circular orbits around their common center of mass. This balance between gravitational forces and the stars' inertia maintains the separation between them.

Part (b)

To determine the mass of each star, we can use Kepler's third law and Newton's version of it for two bodies orbiting each other, which states:

[tex]T^2 = (4*pi^2*r^3)/(G*(m1+m2))[/tex]

Since m1 = m2 (equal masses), we can solve for a single mass. By substituting the given values (radius and period), we can calculate the mass of each star.

(a) The two stars do not crash into each other because they are in a stable orbit, rotating around their common center of mass (barycenter). (b) The mass of each star must be approximately [tex]9.61 \times 10^{11} \, \text{kg}[/tex].

To address the questions about the two stars, we need to use concepts from orbital mechanics and Newtonian gravity. Here's the step-by-step approach:

(a) Why don’t the two stars crash into one another due to the gravitational force between them?

The two stars do not crash into each other because they are in a stable orbit, rotating around their common center of mass (barycenter). The gravitational force between the stars provides the necessary centripetal force to keep them in this orbit.

(b) What must be the mass of each star?

To find the mass of each star, we can use the following relationship between gravitational force and centripetal force. First, let's summarize the given data:

- Distance between the stars, [tex]d = 8.0 \times 10^{10} \, \text{m}[/tex]

- Period of revolution,  T = 12.6 yr

Convert the period  T  into seconds:

[tex]T = 12.6 \, \text{yr} \times \left(365.25 \, \text{days/yr}\right) \times \left(24 \, \text{hours/day}\right) \times \left(3600 \, \text{seconds/hour}\right) \\\\T \approx 12.6 \times 365.25 \times 24 \times 3600 \, \text{s} \approx 3.97 \times 10^8 \, \text{s}[/tex]

Each star rotates around the common center of mass with a radius  r  that is half the distance between the stars, so:

[tex]r = \frac{d}{2} = \frac{8.0 \times 10^{10} \, \text{m}}{2} = 4.0 \times 10^{10} \, \text{m}[/tex]

The gravitational force  F  providing the centripetal force [tex]F_c[/tex] can be equated as follows:

F =  [tex]F_c[/tex]

The gravitational force between the two stars is given by Newton's law of gravitation:

[tex]F = \frac{G m_1 m_2}{d^2}[/tex]

Since the stars have equal mass ( m₁ = m₂ = m ):

[tex]F = \frac{G m^2}{d^2}[/tex]

The centripetal force required to keep each star in orbit is:

[tex]F_c = m \frac{v^2}{r}[/tex]

Where  v  is the orbital velocity of each star. The orbital velocity  v  can be related to the period  T  and the radius r by:

[tex]v = \frac{2 \pi r}{T}[/tex]

Now substituting  v  into the centripetal force equation:

[tex]F_c = m \left( \frac{2 \pi r}{T} \right)^2 \frac{1}{r} \\\\F_c = m \frac{4 \pi^2 r^2}{T^2 r} \\\\F_c = m \frac{4 \pi^2 r}{T^2}[/tex]

Since  F =  [tex]F_c[/tex] :

[tex]\frac{G m^2}{d^2} = m \frac{4 \pi^2 r}{T^2}[/tex]

We can cancel one  m  from each side:

[tex]\frac{G m}{d^2} = \frac{4 \pi^2 r}{T^2}[/tex]

Solving for  m :

[tex]m = \frac{4 \pi^2 r d^2}{G T^2}[/tex]

We know:

[tex]r = 4.0 \times 10^{10} \, \text{m} \\\\d = 8.0 \times 10^{10} \, \text{m} \\\\G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2} \\\\T = 3.97 \times 10^8 \, \text{s}[/tex]

Substitute these values in:

[tex]m = \frac{4 \pi^2 (4.0 \times 10^{10} \, \text{m}) (8.0 \times 10^{10} \, \text{m})^2}{(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2 \cdot \text{kg}^{-2}) (3.97 \times 10^8 \, \text{s})^2}[/tex]

First calculate [tex]r \cdot d^2[/tex]:

[tex]r \cdot d^2 = 4.0 \times 10^{10} \, \text{m} \times (8.0 \times 10^{10} \, \text{m})^2 = 4.0 \times 10^{10} \, \text{m} \times 64.0 \times 10^{20} \, \text{m}^2 = 256.0 \times 10^{30} \, \text{m}^3[/tex]

Then calculate  T² :

[tex]T^2 = (3.97 \times 10^8 \, \text{s})^2 = 15.76 \times 10^{16} \, \text{s}^2[/tex]

Now, plug in the values:

[tex]m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{6.674 \times 10^{-11} \times 15.76 \times 10^{16}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{1.05136 \times 10^6 \times 10^{16}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{30}}{1.05136 \times 10^{22}} \\\\m = \frac{4 \pi^2 \times 256.0 \times 10^{8}}{1.05136} \\\\m = \frac{4 \times 9.8696 \times 256.0 \times 10^{8}}{1.05136} \\\\m = \frac{10102.4 \times 10^{8}}{1.05136} \\\\m \approx 9.61 \times 10^{11} \, \text{kg}[/tex]

Thus, the mass of each star must be approximately [tex]9.61 \times 10^{11} \, \text{kg}[/tex].

A 15.0 lbs block rests on the floor. Then, the force exerted on the block is in the upward direction to counterbalance the Earth's force on the block. If a rope is tied to it, then the force from the floor and the rope together balance the weight.

A force is an external agent which is capable of changing an object's state of rest or motion. Force has both the magnitude and the direction, hence it is a vector quantity. The direction towards which this force is applied is known as the direction of the force, and the application of force is the point at which the force is being applied.

The force exerted on the block by the floor is 15.0lb upwards, to counterbalance the Earth’s force on the block. If a rope is tied to the block, and is running vertically over a pulley, and other object of 10lb is attached to it, then a force of 5.00lb towards upward direction is exerted by the floor on the block. The Earth pulling it down with 15.0lb and the rope pulling it up with a force of 10.0lb. The forces from the floor and rope together balance the weight of objects.

If the 10 lb object is replaced with a 20lb object, then the force exerted by the floor on the 15lb block is 0, because the block now accelerates up away from the floor.

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Answer: The weight of the cannonball on Earth is 200 N

Explanation:

Weight is defined as the force exerted by the body on any surface. It is also defined as the product of mass of the body multiplied by the acceleration due to gravity.

Mathematically,

[tex]W=mg[/tex]

where,

W = weight of the cannonball

m = mass of the cannonball = 20 kg

g = acceleration due to gravity on Earth = [tex]10m/s^2[/tex]

Putting values in above equation, we get:

[tex]W=20kg\times 10m/s^2=200N[/tex]

Hence, the weight of the cannonball on Earth is 200 N