why is the moment of inertia of a hoop that has a mass m and a radius r greater than the moment of inertia of a disk that has the same mass and radius? why is the moment of inertia of a spherical shell that has a mass m and a radius r greater than that of a solid sphere that has the same mass and radius?

On conservation of momentum:

(a) The cart goes in the opposite direction as the man does.

(b) With a speed of 0.07 m/s, the cart proceeds in the opposite direction as the guy.

(c) the cart starts moving in the forward direction with the same velocity of 0.07 m/s.

(a) As per the conservation of momentum, the total momentum of the system is conserved. Initially, the system was at rest, but when the man starts walking towards the other end, he gains some momentum in the forward direction, which the cart has to compensate for. So, the cart moves in the opposite direction to that of the man's motion.

(b) Assume that the man moves a distance of 10 m, i.e., the length of the cart.

Therefore, the total distance covered by the man is 20 m (10 m forward and 10 m backward).

The momentum gained by the man while walking forward is (70 kg) x (1.0 m/s) = 70 kg m/s.

As the total momentum of the system is conserved, the cart gains an equal and opposite momentum of -70 kg m/s.

The mass of the cart is 1,000 kg, so its velocity can be calculated using the conservation of momentum formula:

Total initial momentum = Total final momentum

0 = (70 kg) x (1.0 m/s) + (1,000 kg) x V

V = -0.07 m/s

So, the cart moves in the opposite direction to that of the man's motion with a speed of 0.07 m/s.

(c) When the guy comes to a complete halt at the back of the automobile, he loses the momentum he got while going forward, and the cart obtains equal and opposite motion. As a result, the cart begins going ahead at the same velocity of 0.07 m/s.

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Answer:

Explanation:

operantly conditioned; positively reinforced

The final pressure, when the volume changes to 2.50 L at constant temperature, is 11.25 atm.

To solve this problem, we can use Boyle's Law, which states that the pressure and volume of a gas are inversely proportional at constant temperature.

According to Boyle's Law:

P₁V₁ = P₂V₂

Where P₁ and V₁ are the initial pressure and volume, and P₂ and V₂ are the final pressure and volume.

Given:

P₁ = 2.50 atm

V₁ = 4.50 L

V₂ = 2.50 L

Let's substitute the values into the equation and solve for P₂:

P₁V₁ = P₂V₂

(2.50 atm)(4.50 L) = P₂(2.50 L)

(11.25 atm·L) = (2.50 L)P₂

Dividing both sides of the equation by 2.50 L:

11.25 atm = P₂

Therefore, the final pressure, when the volume changes to 2.50 L at constant temperature, is 11.25 atm.

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The sun's mass is about 333,000 times that of the Earth.

The mass of the sun is approximately 1.989 x 10^30 kg, while the mass of the Earth is about 5.972 x 10^24 kg. To find the ratio of the sun's mass to Earth's mass, you can divide the mass of the sun by the mass of the Earth:
(1.989 x 10^30 kg) / (5.972 x 10^24 kg) ≈ 333,000

Therefore, the sun's mass is roughly 333,000 times greater than the Earth's mass. The significant difference in mass between the sun and the Earth plays a crucial role in the gravitational forces within our solar system, keeping the Earth and other planets in orbit around the sun.

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The force that causes a car to follow a circular path when going around a curve on a horizontal road at a constant speed is the friction force from the road (Option A). This force is essential for the car to overcome the tendency to move in a straight line and maintain its curved trajectory.

When a car goes around a circular curve, it experiences a centripetal acceleration directed towards the center of the curve. According to Newton's second law of motion, F = ma, there must be a net force acting on the car to produce this acceleration. In this case, the friction force between the car's tires and the road provides the necessary centripetal force.

The car has a tendency to move in a straight line due to its inertia, as described by Newton's first law. However, the curved path requires a force to redirect its motion.

As the car turns, the tires exert a friction force on the road in the opposite direction of the car's motion. This force arises from the interaction between the microscopic irregularities on the tire and the road surface.

The friction force acts as the centripetal force, directed towards the center of the circular path. It enables the car to change its direction and continually adjust its trajectory to follow the curve.

The normal force from the road (Option B) and gravity (Option C) are present but not directly responsible for the car's circular motion. The normal force acts perpendicular to the road's surface, counteracting the weight of the car and preventing it from sinking into the road.

Option D, which suggests that no force is causing the car to follow the circular path, is incorrect. Even though the car is traveling at a constant speed and has no linear acceleration, it experiences a centripetal acceleration that requires a force (friction) to maintain the circular trajectory.

In conclusion, the correct answer is A) the friction force from the road, which provides the necessary centripetal force for the car to follow the circular path.

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The magnitude of the point charge is approximately 2.7 x 10⁻⁸ coulombs.

We can use Coulomb's law to solve for the magnitude of the point charge. Coulomb's law states that the electric field, E, at a distance r from a point charge, q, is given by: E = k * (q / r²)

where k is Coulomb's constant, which is approximately equal to 8.99 x 10⁹ N * m² / C².

In this case, we are given the electric field magnitude, E = 2.3 n/C, and the distance from the point charge, r = 56 cm = 0.56 m. We can rearrange Coulomb's law to solve for the magnitude of the point charge, q: q = E * r² / k

Substituting the given values, we get: q = (2.3 n/C) * (0.56 m)² / (8.99 x 10⁹ N * m² / C²)

q = 2.7 x 10⁻⁸ C

Therefore, the magnitude of the point charge is approximately 2.7 x 10⁻⁸ coulombs.

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Answer:

B

Explanation:

Wild chimpanzees fail to use certain aspects of human language in their systems of calls about predators, such as syntax and grammar. They also do not have the ability to create new words or abstract concepts, which are key components of human language.

When bonobos learn human sign-language or a pictogram language, they may be weak on certain aspects of human language such as syntax and grammar, as well as the ability to understand figurative language, metaphors, and idioms. They may also struggle with understanding complex sentences and communicating complex ideas. However, with proper training and practice, bonobos can develop impressive communication skills using these artificial languages.
Hi! Wild chimpanzees fail to use certain aspects of human language in their systems of calls about predators, such as syntax, grammar, and complex vocabulary. Additionally, they lack the ability to produce and comprehend a wide range of sounds or symbols that represent specific concepts.

When bonobos learn human sign language or a pictogram language, they tend to be weak in areas such as grammar, syntax, and the ability to create complex sentences. They may also struggle with understanding idiomatic expressions, metaphors, and other abstract language features.

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Answer:  14 inches.

Explanation: The radius of this circle is 7 inches, so the diameter is 2(7) = 14 inches. The diameter is 14 inches.

Answer:

1.5 N

Explanation:

You've left us to guess what the question is. I will Assume it is what's the force?

Givens

m = 3 kg

vi = 1.5 m/s

vf = 4 m/s

t = 5 seconds

Formula

F = m * (vf - vi)/t

Solution

F = 3 * (4 - 1.5) / 5

F = 1.5 N

Answer:

Hiiiiiiii........where r u frm??

Answer:

D. P=1.2 kw, T=6kwh

Explanation:

P = A × V

P = 120×10

= 1200 watts or 1.2 kw

T = Watts × hr

1000

= 1200 × 5

1000

= 6 kwh

The highest order dark fringe for a 561 nm light incident on a 1420 nm wide slit is the 3rd order.

Diffraction occurs when light passes through a narrow opening or slit, causing the wave to bend and interfere with itself. The pattern of bright and dark fringes produced by this interference is called a diffraction pattern. The position of these fringes can be determined using the equation d sin θ = mλ, where d is the width of the slit, θ is the angle of diffraction, m is the order of the fringe, and λ is the wavelength of the light.

Using this equation, we can calculate that the 3rd order dark fringe corresponds to an angle of approximately 5.68 degrees for a 561 nm light incident on a 1420 nm wide slit. Therefore, the highest order dark fringe in this situation is the 3rd order.

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The line of evidence that is not consistent with the theory of evolution of the horse is: "The fossils of the limbs are different from each other, which indicates these organisms are not related and evolved separately from each other."

The theory of evolution suggests that organisms share a common ancestor and have evolved through gradual changes over time. Therefore, the presence of similar structures and traits among fossilized limbs is consistent with the theory, indicating that these organisms are related and evolved from a common ancestor. If the fossils of the limbs were significantly different from each other, it would suggest that these organisms evolved separately and do not share a common ancestry. However, this would contradict the theory of evolution, which states that species are connected through common ancestors and have undergone a process of gradual change.

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Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

We need to verify the thrid equation of kinematics here

\(\\ \tt\longmapsto v^2-u^2=2as\)

\(\\ \tt\longmapsto 20^2=2(-8)s\)

\(\\ \tt\longmapsto 400=-16s\)

\(\\ \tt\longmapsto s=|400/-16|\)

\(\\ \tt\longmapsto s=25m\)

The squirrel has a good luck ,Car gets stopped just 1m away from the squirrel .

Explanation:

According to the Question

A squirrel runs out in front of a car moving at +20m/s ( initial speed u) . The driver slams on the brakes and begins decreasing at -8m/s² ( here the acceleration is acting in opposite directions of motion a )

When Squirrel was 26 metres ( Distance from the car initially )

Using Kinematic Equation

On substituting the value we obtain

➥ 0² = 20² + 2 × -8 × s

➥ 0 = 400 - 16 × s

➥ -400 = -16 × s

➥ 400/16 = s

➥ 25 = s

So, the squirrel is 25 metres from the car .

True, it is possible for a fault to have a north-south (N-S) strike and an eastward dip.

Faults can have various orientations and dips depending on the tectonic forces and geological conditions in a particular region. The strike refers to the direction of the fault line on the Earth's surface, while the dip indicates the angle at which the fault plane is inclined from the horizontal. Therefore, a fault can have any combination of strike and dip, including an N-S strike and an eastward dip.

Faults are geological fractures where movement has occurred, and their orientations can vary. A fault with a north-south (N-S) strike means it extends in that direction horizontally. The dip refers to the angle of the fault plane from the horizontal. So, it is possible for a fault to have an N-S strike and an eastward dip.

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Answer:

modern world

Explanation:

The acceleration of the particle after 1 minute is \(18 units/min^2.\)

To find the acceleration, we need to differentiate the position function twice with respect to time.

Given:

\(s(t) = 2t^3 + 3t^2 + t\)

To find the velocity, differentiate s(t) with respect to t:

\(v(t) = d/dt [2t^3 + 3t^2 + t]\)

\(v(t) = 6t^2 + 6t + 1\)

To find the acceleration, differentiate v(t) with respect to t:

\(a(t) = d/dt [6t^2 + 6t + 1]\)

\(a(t) = 12t + 6\)

Now, substitute t = 1 to find the acceleration after 1 minute:

\(a(1) = 12(1) + 6\)

\(= 12 + 6\)

\(= 18\)

Therefore, the acceleration after 1 minute is \(18 units/min^2\)

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Answer:

the temperature increas

Answer:

See the answers below.

Explanation:

a) KEf, This term refers to the final kinetic energy i.e. when the velocity has reached 27 [m/s].

To calculate this energy we must use the following equation that relates velocity and mass to kinetic energy.

\(E_{kin}=\frac{1}{2} * m*v^{2}\)

where:

m = mass = 2.2*10³[kg]

v = velocity = 27 [m/s]

\(E_{kin}=\frac{1}{2} *2.2*10^{3}*(27)^{2}\\E_{kin}=801900[J]\)

b)

Power is defined as the amount of energy or work done over a certain time, it can be calculated by means of the following equation.

\(P=E_{kin}/t\)

where:

t = time = 2.4 [s]

P = Power [W]

Now replacing:

\(P=801900/2.4\\P=334125[W]\\Now in Hp\\334125[w]*\frac{1hp}{745w} \\= 448.5[hp]\)

To find the voltage, δv=vd−vc, between points c and d, we need to measure the potential difference between these two points. This can be done using a voltmeter, which is a device that measures voltage.

Assuming that we have a circuit with points c and d, we can connect the voltmeter between these two points to measure the potential difference. The positive (+) lead of the voltmeter should be connected to point d, and the negative (-) lead should be connected to point c.

The voltage measured by the voltmeter will be the potential difference between points c and d, which is equal to δv=vd−vc. Therefore, we can read the voltage value directly from the voltmeter to find the potential difference between points c and d.

Note that the voltage value will depend on the specific circuit and the voltages at points c and d relative to a common reference point (such as ground).

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Answer:

This is just simple math. Your answer would be 26.7 m/s

Explanation:

Answer:

4 ohms

Explanation:

Current = Voltage/resistance

2 = 8/R

2R = 8

R = 4

The shear and moment diagrams can be determined by evaluating the shear force and moment equations for a given function.

Shear force is a type of force that is created when two objects are pushed against each other in opposite directions.

The shear force equation is defined as the integral of the load over the length of the beam, while the moment equation is defined as the integral of the shear force over the length of the beam.

For example, if the load on the beam is a constant, the shear equation would be:
V(x) = F
Where F is the constant applied load.
The moment equation would be:
M(x) = Fx
Where F is the constant applied load and x is the distance along the beam.

The shear and moment diagrams can then be drawn by plotting the shear and moment equations as a function of x. The shear diagram will simply be a horizontal line at the value of the constant load, while the moment diagram will be a straight line that increases linearly with x.

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Answer:

c and d

Explanation:

The wave shown in the figure is Sound waves. The sound wave is a mechanical wave. Hence, options C and D are correct.

Sound is traveled by means of vibrations of particles in the medium. Sound waves require a medium to propagate. The energy propagation by means of adiabatic loading and unloading is called sound or acoustic waves.

The sound waves produce compression and rarefaction in the air. The sound waves are called Longitudinal waves because the direction of propagation is parallel to the vibration of particles.

Sound waves are considered only as waves because they exhibit a wave nature. Light has both properties of wave and particle. Light is an electromagnetic wave and it is transverse in nature.

The sound waves require a medium like air, water, solids, etc for propagation. Hence it is called Mechanical waves. Light waves do not require a medium to propagate. From the given wave pattern,  it exhibits the property of sound waves.

Hence, the correct option is C) the wave is a mechanical wave, and D) the wave could be transmitting sound.

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Answer: Echolocation

Explanation:

a) The push required to give the crate an acceleration of 1.13 m/s^2 is 360.476 N.

b) If we maintain the push of 231 N, the crate would slow down with an acceleration of 1.734 m/s^2 in the opposite direction.

a) To give the crate an acceleration of 1.13 m/s^2, we need to find the net force required.

The force required to overcome static friction and just start the crate moving is given by

F_static = friction coefficient × N

where N is the normal force exerted by the floor on the crate. Since the crate is on a horizontal floor, the normal force is equal to its weight

N = m×g

where m is the mass of the crate and g is the acceleration due to gravity.

N = 45.2 kg × 9.81 m/s^2 = 443.812 N

The force required to overcome static friction is

F_static = friction coefficient × N

We know that the force required to just start the crate moving is 310 N, so the friction coefficient is

friction coefficient = F_static / N = 310 N / 443.812 N = 0.698

Once the crate is moving, the force required to keep it moving at a constant speed is equal to the force of kinetic friction

F_kinetic = friction coefficient × N = 0.698 × 443.812 N = 309.4 N

To give the crate an acceleration of 1.13 m/s^2, we need a net force of

F_net = m×a = 45.2 kg × 1.13 m/s^2 = 51.076 N

So, the force required to give the crate an acceleration of 1.13 m/s^2 is

F_push = F_net + F_kinetic = 51.076 N + 309.4 N = 360.476 N

b) If we maintain the push at 231 N, the net force on the crate would be

F_net = F_push - F_kinetic = 231 N - 309.4 N = -78.4 N

The negative sign indicates that the net force is in the opposite direction to the motion of the crate, so it would slow down with an acceleration of

a = F_net / m = -78.4 N / 45.2 kg = -1.734 m/s^2

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The given question is incomplete, the complete question is:

A crate of 45.2-kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 310N . After that you must reduce your push to 231N to keep it moving at a steady 25.2cm/s .

a) What push must you exert to give it an acceleration of 1.13m/s^2 ? b) What would its acceleration be if you maintained the push?

Hi there!

Assuming the rod is pulled back 90° (π/2 radians), we can use angular kinematics to solve.

Using the equation:

\(\omega_f^2= \omega_i^2 + 2\alpha \theta\)

\(\omega_f\) = final angular velocity (? rad/sec)

\(\omega_i\) = initial angular velocity (0 rad/sec)

\(\alpha\) = angular acceleration (rad/sec², must solve for)

\(\theta\) = angular displacement (π/2 rad)

Recall Newton Second Law's rotational equivalent:
\(\Sigma \tau = I\alpha\)

τ = Torque (Nm)

I = Moment of inertia (kgm²)
α = angular acceleration (rad/sec²)

Setting the pivot point at the end, and knowing that the force of gravity works at an object's center of mass (1.5m from the rod's end), we can solve for the angular acceleration.

\(\tau= F \times r\\\\\tau = W \times r = 10(9.8) * 1.5 = 147 Nm\)

Now, solve for the moment of inertia given I = 1/3ML² at one of its ends:
\(I = \frac{1}{3}(10)(3^2) = 30 kgm^2\)

Solve for the angular acceleration:
\(147 = 30\alpha\\\\\alpha = 4.9 \frac{rad}{sec^2}\)

Now, we can use the kinematic equation to solve.

\(\omega_f^2 = 0 + 2(4.9)(\frac{\pi}{2}})\\\\\omega_f = \sqrt{2(4.9)(\frac{\pi}{2})} = \boxed{3.923 \frac{rad}{sec}}\)