A hockey puck with a mass of 0.171 kg slides over the ice. The puck initially slides with a speed of 5.65 m/s, but it comes to a rough patch in the ice which slows it down to a speed of 3.05 m/s.

How much energy is dissipated as the puck slides over the rough patch?

a) Minimum speed required for the riders so that they stay pinned against the inside of the drum is 18m/s.

b) Angular velocity of the drum at this speed having diameter of the drum is 10 m is 3.6 rad/s

Friction is a resistance to motion of the object. for example, when a body slides on horizontal surface in positive x direction, it has friction in negative x direction and that measure of friction is a frictional force.

frictional force is directly proportional to the Normal(N).

i.e. [tex]F_{fri}[/tex] ∝ N

[tex]F_{fri}[/tex] = μN

where μ is called as coefficient of the friction. It is a dimensionless quantity.

When a body is kept on horizontal surface, its normal will be straight upward which is reaction of mg. i.e. N=mg.

Given,

Diameter of the drum D = 10m , Radius r = 5m

Coefficient of static friction = 0.15

a) To stay everyone pinned against the wall of drum. Frictional Force must be equal to weight mg which are opposite to each other.

μN = mg ........1)

Centrifugal acceleration = Normal

mv²÷r = N

With this equation 1 becomes

μ[tex]\frac{mv^{2} }{r}[/tex] = mg

v² = rg÷μ

v² = 5m*9.8m/s ÷ 0.15

v² = 326.6

v= [tex]\sqrt{326.6}[/tex] = 18.07 ~ 18 m/s

Hence minimum speed required for the riders is 18m/s.

b) for angular velocity of the drum, V=rω

ω = v÷r

ω = 18m/s ÷ 5m

ω = 3.6 rad/s

hence angular velocity of the drum at 18m/s speed is 3.6 rad/s

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Answer:

Without the table and diagram, it's difficult to provide a specific answer. However, in general, the gravitational attraction between two objects depends on their masses and the distance between them. The force of attraction is stronger when the masses are greater and when the objects are closer together.

Based on this understanding, we can evaluate the claims made by each student:

Student W claims that the probe is pulled harder by the planet with the greater mass when halfway between the planets. This claim is partially supported by the evidence, since the gravitational force is stronger when the masses are greater. However, it doesn't take into account the distance between the planets and the probe.

Student X claims that the probe is pulled harder by the planet nearest the Sun when halfway between the planets. This claim is not supported by the evidence, since the distance between the probe and each planet is not specified.

Student Y claims that the probe is pulled harder by the planet with the greater mass anywhere between the planets. This claim is partially supported by the evidence, since the gravitational force is stronger when the masses are greater. However, it doesn't take into account the distance between the planets and the probe.

Student Z claims that the probe is pulled harder by the planet nearest the Sun anywhere between the planets. This claim is not supported by the evidence, since the distance between the probe and each planet is not specified.

Overall, Student W and Student Y made claims that are partially supported by the evidence, but neither claim takes into account the distance between the planets and the probe. Therefore, it's difficult to determine which claim is best supported by the evidence without more information.

Explanation:

Answer:

Explanation:

a) transverse

b)longitudinal waves

Answer:

To solve this problem, we need to use the law of conservation of momentum, which states that the total momentum before a collision is equal to the total momentum after the collision.

The momentum of an object is given by its mass multiplied by its velocity, so we can calculate the initial momentum of each ball before the collision:

- The northbound ball has a momentum of 535g * 12.4 m/s = 6644 g*m/s north

- The southbound ball has a momentum of 725g * (-6.4 m/s) = -4640 g*m/s north (note that the negative sign indicates southward direction)

Adding these momenta together, we get a total momentum of 6644 g*m/s - 4640 g*m/s = 2004 g*m/s north.

After the collision, the two clay balls stick together and move as a single unit. Let's call the mass of the combined unit "M" and its velocity "v". By conservation of momentum, we know that the total momentum of the combined unit after the collision must be the same as the total momentum before:

M * v = 2004 g*m/s north

To solve for v, we need to figure out the mass of the combined unit. This is simply the sum of the masses of the two original balls:

M = 535g + 725g = 1260g

Substituting this into the equation above, we get:

1260g * v = 2004 g*m/s north

Solving for v, we get:

v = 1.59 m/s north

Therefore, the combined unit moves 1.59 m/s north after the collision.

However, the answer choices given in the problem are in meters per second, not meters per second north/south. To convert the answer, we need to add a direction. Recall that the northbound ball had a positive velocity and the southbound ball had a negative velocity. Since the combined unit is moving northward, we know its velocity must be positive.

Therefore, the final answer is 1.59 m/s north, which corresponds to answer choice C.

Answer:

To calculate the work being done, we need to use the formula:

work = force x distance x cos(theta)

where force is in newtons, distance is in meters, and theta is the angle between the direction of the force and the direction of the movement.

In this case, the force is the weight of the leaf blower, which is 1600 N, the distance is 6 meters, and the angle between the force and the movement is 0 degrees (since Kyle is lifting the leaf blower straight up). So we have:

work = 1600 N x 6 m x cos(0°)

work = 9600 J

Therefore, the joules of work being put out by Kyle are 9600 J.

Explanation:

Specific heat is a measure of how much energy it takes to raise the temperature of a substance.

Temperature is defined as the measurement of degree of amount of hotness or coldness of a body.

Here,

Specific heat is a measure of how much energy it takes to raise the temperature of a substance. More clearly, specific heat is the amount heat required to raise the temperature of unit mass through one degree.

If an amount of heat Q is given to a body of mass m and ΔT is the rise in temperature. Then specific heat capacity,

      C = Q/mΔT

Its unit is Jkg⁻¹K⁻¹

Specific heat of a substance is a constant but it changes slightly with change in temperature.

The rise in temperature is small for body having large specific heat.

Hence,

Specific heat is a measure of how much energy it takes to raise the temperature of a substance.

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Answer: 12 meters

Explanation:

the rate is 4 meters per second
the rock rolled for 3 seconds
4 x 3 = 12

Answer:

The wavelength of the photon can be calculated using the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 ✕ 10^−34 J s), c is the speed of light (2.998 ✕ 10^8 m/s), and λ is the wavelength.

λ = hc/E = (6.626 ✕ 10^−34 J s) (2.998 ✕ 10^8 m/s) / (3.02 ✕ 10^−19 J) = 656.4 nm

Explanation:

Answer:

b

Explanation:

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Answer: 20

Explanation:

The area of the bone is 0.785 m². Then the maximum force that can be exerted on the femur bone if the shear stress is 9 × 10⁷ N/m² is 70650 kN.

The Young's modulus of a material is the ratio of its stress to strain. Where stress is the force per unit area and strain be the ratio of change in length to the original length.

given stress s = 9 × 10⁷ N/m²

diameter of the bone d = 9 cm = 0.09 m.

then area = π d²/4

a = 3.14 × (0.09 m )²/4 = 0.785 m².

Stress = maximum force/area

then Fmax  = stress × area

Fmax = 9 × 10⁷ N/m²  × 0.785 m²

         = 70650 kN.

Therefore, the maximum force that can be exerted to the bone is 70650 kN.

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Answer:

In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next. This means that the sound wave travels over four times faster than it would in air, but it takes a lot of energy to start the vibration.

Explanation:

Answer:

slower as it is more dense in water

Explanation:

Answer:

To calculate the resistance of the microwave, we can use Ohm's Law, which states that:

resistance = voltage / current

Substituting the given values into this equation, we get:

resistance = 5V / 0.3A

resistance = 16.67 ohms

Therefore, the resistance of the microwave with 5V and a current of 300mA is 16.67 ohms.

The radius of a water droplet suspended in an electric field can be calculated using the following formula:

r = (3qE/4πρg)^(1/2)

where r is the radius of the droplet, q is the charge on the droplet, E is the strength of the electric field, ρ is the density of the droplet material (assumed to be that of water, which is 1000 kg/m^3), and g is the acceleration due to gravity.

In this case, q = -1.602 × 10^-19 C (the charge on an electron), E = 300 V/m, ρ = 1000 kg/m^3, and g = 9.81 m/s^2.

Plugging in these values, we get:

r = (3(-1.602 × 10^-19 C)(300 V/m)/(4π(1000 kg/m^3)(9.81 m/s^2)))^(1/2)

= 1.83 × 10^-5 m

Therefore, the radius of the water droplet is approximately 1.83 × 10^-5 meters.

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Answer:

5000 km

Explanation:

Sure, to calculate the diameter of Vesta, we can use the fact that Vesta's diameter is equal to the diameter of Mercury, which is 5000 km.

Therefore, the diameter of Vesta is also 5000 km.

We can show the working by using the following formula:

Diameter of Vesta = Diameter of Mercury

Diameter of Vesta = 5000 km

yes they would balance because they have the same weight

Explanation:

hope this helps (:

Since the two children have the same weight, they have to sit at equal distances from the pivot of the seesaw.  Then they will balance.

Young’s modulus for each of these five materials from largest to smallest are mentioned below.

What is Young’s modulus?

Many substances lack linearity and elasticity after very minor deformation. Only materials that are linearly elastic are subject to the constant Young's modulus. In this case, the ratio of stress to strain, which corresponds to the material's stress, determines the Young's modulus.

What is data?

Facts such as numbers, words, measurements, observations, or even simple descriptions of objects are grouped together as data. Both qualitative and quantitative data are possible. Qualitative data is information that is descriptive (describes something), whereas discrete data can only take particular values (like whole numbers).

Therefore, Young’s modulus for each of these five materials from largest to smallest are mentioned above.

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Answer:

The frequency of a mass vibrating on a string describes the number of complete cycles of vibration that occur per unit of time, usually measured in hertz (Hz).

Explanation:

The wavelength of maximum intensity for a celestial body with a temperature of 50000 K is approximately 577 nm.

To find the wavelength of maximum intensity for a celestial body with a temperature of 50000 K, we can use Wien's displacement law, which relates the temperature of an object to the wavelength at which it emits the most radiation. The law is given by:

λ_max = b / T

where λ_max is the wavelength of maximum intensity, b is Wien's displacement constant (2.898 × 10⁻³ m·K), and T is the temperature in Kelvin.

We can convert the temperature of 50000 K to Kelvin by adding 273.15 to get:

T = 50000 K + 273.15 = 50273.15 K

Plugging in the values, we get:

λ_max = (2.898 × 10⁻³ m·K) / 50273.15 K

Simplifying, we get:

λ_max = 5.77 × 10⁻⁸ meters

To convert meters to nanometers, we can multiply by 10⁹:

λ_max = 5.77 × 10⁻⁸ meters × 10⁹ nm/m = 577 nm

Therefore, the wavelength of maximum intensity for a celestial body with a temperature of 50000 K is approximately 577 nm.

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Answer:

B. 1/2mv^2 + 1/2Iw^2

Explanation:

Answer:

Explanation:

Power=work done/time

work done=force* displacement

force=mass*acceleration due to gravity

Therefore, power= mass*acceleration due to gravity*displacement/time

20*9.8*1/1=196watts

The power loss in the bearing is 1.36 W shaft of 100 mm diameter rotates at 120 rad/s in a bearing 150 mm long.

The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.

The viscosity of a fluid is a measure of its resistance to flow. A fluid with a high viscosity resists movement because its molecular structure causes a lot of internal friction.

To calculate the power loss in the bearing, we can use the formula:

P = F × V

To calculate the frictional force, we can use the formula:

F = μ × A × P

Where μ is the coefficient of friction, A is the area of the bearing, and P is the pressure of the lubricant.

First, we need to calculate the pressure of the lubricant:

P = μ × Viscosity × (V/D)

Where D is the diameter of the shaft.

P = (0.02) × 20 × (120/0.1) = 480 N/[tex]m^2[/tex]

Next, we can calculate the area of the bearing:

A = π × (D/2)^2 × L

Where L is the length of the bearing.

A = π × (0.1/2)^2 × 0.15 = 0.001178 [tex]m^2[/tex]

Now we can calculate the frictional force:

F = μ × A × P = (0.02) × 0.001178 × 480 = 0.0113 N

Finally, we can calculate the power loss:

P = F × V = 0.0113 × 120 = 1.36 W

Therefore, the power loss in the bearing is 1.36 W.

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The magnitude of the electrical field would be 1.5x10^3 N/C.

The electrostatic force experienced by a charged particle in an electric field is given by the formula:

F = qE

Where F is the electrostatic force, q is the charge of the particle, and E is the electric field strength.

In this case, the electrostatic force acting on the particle is 3.0x10^-6 N and the charge of the particle is -2.0x10^-9 C. So we have:

3.0x10^-6 N = (-2.0x10^-9 C)E

Solving for E, we get:

E = (3.0x10^-6 N) / (-2.0x10^-9 C)

E = -1.5x10^3 N/C

Since the electric field is a vector quantity, its magnitude is always positive. Therefore, the magnitude of the electric field in this case is:

|E| = 1.5x10^3 N/C

So the magnitude of the electric field is 1.5x10^3 N/C, directed downward.

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The solutions for questions A and B are

Generally, Tension is a physical force that pulls on an object, often tending to stretch it. It is typically measured in units of force per area, such as newtons per square meter (N/m2). Tension is an important concept in mechanics, physics, engineering, and other fields.

(a) T_1=m_1(g-a)

T_1=4.81 (9.8-3.94)

T_1=28.1866N

(b) In freefall

a=g

T_1=m_1(g-g)

T_1=0 N

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Answer:

20.5 m/s

Explanation:

Answer:

a) On the Moon, where the acceleration due to gravity is 0.258g:

First, we need to find the acceleration due to gravity on the Moon:

g_Moon = 0.258g_Earth

g_Moon = 0.258(12.3 m/s^2)

g_Moon = 3.17 m/s^2

Now we can use the range formula for projectile motion to find the distance he could jump:

R = (v^2/g) sin(2θ)

Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:

R = (v^2/g) sin(^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)

R = 0.191 m

Therefore, he could jump approximately 0.191 m on the Moon.

!

a) On the Moon, where the acceleration due to gravity is 0.258g:

First, we need to find the acceleration due to gravity on the Moon:

g_Moon = 0.258g_Earth

g_Moon = 0.258(12.3 m/s^2)

g_Moon = 3.17 m/s^2

Now we can use the range formula for projectile motion to find the distance he could jump:

R = (v^2/g) sin(2θ)

Assuming the same initial velocity and angle of jump, we can rearrange the formula to solve for R:

R = (v^2/g) sin(

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^2/g_Earth) sin(2θ) * (g_Moon/g_Earth)

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.17 m/s^2) / (12.3 m/s^2)

R = 0.191 m

Therefore, he could jump approximately 0.191 m on the Moon.

b) On Mars, where the acceleration due to gravity is 0.293g:

Similarly, we need to find the acceleration due to gravity on Mars:

g_Mars = 0.293g_Earth

g_Mars = 0.293(12.3 m/s^2)

g_Mars = 3.61 m/s^2

Using the same formula and rearrangement as in part a, we can find the distance he could jump on Mars:

R = (1.31 m)^2/ (212.3 m/s^2) * sin(2θ) * (3.61 m/s^2) / (12.3 m/s^2)

R = 0.223 m

Therefore, he could jump approximately 0.223 m on Mars.

Answer:

less than 100-J

Explanation:

The potential energy held in the stretched rubber band is turned into kinetic energy of the rock when it is released, assuming that the rubber band is used to launch a rock from a slingshot.

The mass of the rock and the effectiveness of the slingshot in transmitting the energy from the rubber band to the rock are two elements that affect how much kinetic energy the rock will have. To estimate the kinetic energy, though, we may make certain generalizations.

Assume that no energy is lost as a result of friction or air resistance and that the entire potential energy held in the rubber band is transformed into the kinetic energy of the rock. In this hypothetical situation, the potential energy of the stretched rubber band would be equal to the kinetic energy of the rock.

As a result, the rock will have 100 J of kinetic energy when it exits the slingshot if the rubber band contains 100 J of potential energy. The actual kinetic energy of the rock would be less than 100 J since some energy will be wasted owing to things like friction.

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Answer: 9.8 J

Explanation:

Since the gravitational potential energy of the object is mgh or mass*acceleration due to gravity*initial height, its [tex]U_{g}[/tex] is 9.8 J. Due to the Law of Conservation of Energy, its kinetic energy will also be 9.8 J. This can be seen in the equation [tex]KE_{i}+ PE_{i}= KE_{f} + PE_{f}[/tex]. Since there is no initial kinetic energy and no final potential energy, its initial potential energy is equal to its final kinetic energy.

Answer:

Explanation: In anyway can you get a better picture just a close picture i cant read it