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A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.72 x 10^4 rad/s to an angular speed of 5.42 x 10^4 rad/s. In the process, the bit turns through 1.72 x 10^4 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 8.42 x 10^4 rad/s, starting from rest?

Answers

Answer 1

Answer:

Answer:

The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.

Explanation:

ω1= 1.72x10^4 rad/sec

ω2= 5.42x10^4 rad/sec

ωmax= 8.42x10^4 rad/sec

θ= 1.72x10^4 rad

$\alpha = (w2^(2)-w1^(2) )/(2*(\theta2 - \theta1))$

α=7.67 x10^4 rad/sec²

t= ωmax / α

t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²

t=1.097 sec

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A particle located at the position vector m has a force N acting on it. The torque about the origin is

Answers

Final answer:

The torque about a given origin when a force N is acting on a particle at the position vector m is given by the cross product of the position and force vectors. It's represented by the SI unit Newton-meters, and for multiple particles, the total angular momentum is the vector sum of their individual angular momenta.

Explanation:

The torque about a given origin, when a force N is acting on a particle located at the position vector m, is calculated using the cross product of the position vector and the force vector. This can be written as τ = m x N. The SI unit of torque is Newton-meters (N.m).

As an example, if you apply a force perpendicularly at a distance from a pivot point, you will create a torque relative to that point. Similarly, the torque on a particle is also equal to the moment of inertia about the rotation axis times the angular acceleration.

If we consider multiple particles, the total angular momentum of these particles about the origin is the vector sum of their individual angular momenta. This is calculated by the expression for the angular momentum Ỉ = ŕ x p for each particle, where ŕ is the vector from the origin to the particle and p is the particle's linear momentum.

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Final answer:

The torque on a particle at a position vector m with force N acting on it is calculated by taking the cross-product of the position vector and the force. This principle is the same even in systems with multiple particles. The SI unit of torque is Newton-meters (N·m), which should not be confused with Joules (J).

Explanation:

The torque on a particle located at a position vector m with a force N acting on it is calculated by taking the cross-product of the position vector and the force. In terms of physics, torque (τ) is a measure of the force that can cause an object to rotate about an axis, and it is calculated as the product of the force and the distance from the axis of rotation to the point where force is applied. Hence, the formula for torque is τ = r x F where r is the position vector (or distance from the origin to the point where the force is applied) and F is the force. Remember, this equation gives a vector result with a direction perpendicular to the plane formed by r and F and a magnitude equal to the product of the magnitudes of r and F and the sine of the angle between r and F.

The same principle applies to systems where multiple particles are present. The total angular momentum of the system of particles about a particular point is the vector sum of the individual angular momenta about that point. Torque is the time derivative of angular momentum.

The SI unit for torque is Newton-meters (N·m), which should not be confused with Joules (J), as both have the same base units but represent different physical concepts. In this context, a net force of 40N acting at a distance of 0.800m from the origin would generate a torque of 32 N·m at the origin.

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If you know the distance of an earthquake epicenter from three seismic stations, how can you find the exact location of the epicenter of the earthquake.

Answers

You draw 3 circles around the stations with the size of the circle equal to the distance from the earthquake. Then you simply find where the edge circles all overlap.

Question Part Points Submissions Used A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.30 m/s in 0.812 s. (a) What is the magnitude of the linear impulse experienced by a 62.0-kg passenger in the car during the time the car accelerates? kg · m/s (b) What is the magnitude of the average total force experienced by a 62.0-kg passenger in the car during the time the car accelerates? N

Answers

(a) 328.6 kg m/s

The linear impulse experienced by the passenger in the car is equal to the change in momentum of the passenger:

$I=\Delta p = m\Delta v$

where

m = 62.0 kg is the mass of the passenger

$\Delta v$ is the change in velocity of the car (and the passenger), which is

$\Delta v = 5.30 m/s - 0 = 5.30 m/s$

So, the linear impulse experienced by the passenger is

$I=(62.0 kg)(5.30 m/s)=328.6 kg m/s$

(b) 404.7 N

The linear impulse experienced by the passenger is also equal to the product between the average force and the time interval:

$I=F \Delta t$

where in this case

$I=328.6 kg m/s$ is the linear impulse

$\Delta t = 0.812 s$ is the time during which the force is applied

Solving the equation for F, we find the magnitude of the average force experienced by the passenger:

$F=(I)/(\Delta t)=(328.6 kg m/s)/(0.812 s)=404.7 N$

A mass of 0.14 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.28 m)cos[(8 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass .

Answers

Answer:

The amplitude of oscillation for the oscillating mass is 0.28 m.

Explanation:

Given that,

Mass = 0.14 kg

Equation of simple harmonic motion

$x(t)=(0.28\ m)\cos[(8\ rad/s)t]$ ....(I)

We need to calculate the amplitude

Using general equation of simple harmonic equation

$y=A\omega \cos\omega t$

Compare the equation (I) from general equation

The amplitude is 0.28 m.

Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.

The presence of dwarf galaxies around the Milky Way supports what picture of our galaxy’s formation?

Answers

Answer:

The presence of dwarf galaxies around the Milky Way supports what picture that our galaxy was formed by a coming together or combination of smaller systems

A 0.26-kg stone is held 1.2 m above the top edge of a water well and then dropped into it. The well has a depth of 4.5 m. (a) Relative to the configuration with the stone at the top edge of the well, what is the gravitational potential energy of the stone−Earth system before the stone is released?