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

Answers

Answer 1

Answer:

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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Answer 2

Answer:

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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a force of 35N is exerted over a cylinder with an area of 5m^2. What pressure,in pascals, will be transmitted in the hydraulic system?
A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?
Express the following speeds as a function of the speed of light, c: (a) an automobile speed (93 km/h) (b) the speed of sound (329 m/s) (c) the escape velocity of a rocket from the Earth's surface (12.1 km/s) (d) the orbital speed of the Earth about the Sun (Sun-Earth distance 1.5×108 km).

A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?

Answers

Answer:

part a)

k = 310 N/m

part b)

T = 0.51 s

Explanation:

Part A)

As per work energy theorem we have

Work done by gravity + work done by spring = change in kinetic energy

$mgx - (1)/(2)kx^2 = 0$

$(2.07)(9.8)(0.131) - (1)/(2)k(0.131)^2 = 0$

now we will have

$k = 310 N/m$

Part B)

Time period of oscillation is given as

$T = 2\pi\sqrt{(m)/(k)}$

$T = 2\pi\sqrt{(2.07)/(310)}$

$T = 0.51 s$

Two cars, a Porsche Boxster convertible and a Toyota Scion xB, are traveling at constant speeds in the same direction. Suppose, instead, that the Boxster is initially 170 m behind the Scion. The speed of the Boxster is 24.4 m/s and the speed of the Scion is 18.6 m/s. How much time does it take for the Boxster to catch the Scion

Answers

Answer:

It will take 29.31 seconds for the Boxster to catch the Scion

Explanation:

Given the data in the question;

lets say Toyota Scion xB is car A and Porsche Boxster convertible is B and Toyota Scion xB is car A

the distance travelled by car A is

x = $V_(A)$ × t

where $V_(A)$ is the speed of the car and t is time

the distance travelled by car B before reaching car A will be;

x + x₀ = $V_(B)$ × t

Now lets replace x by $V_(A)$ × t

( $V_(A)$ × t) + x₀ = $V_(B)$ × t

x₀ = ( $V_(B)$ × t) - ( $V_(A)$ × t)

x₀ = t ( $V_(B)$ - $V_(A)$ )

t = x₀ / ( $V_(B)$ - $V_(A)$ )

so we substitute

t = 170 m / (24.4 - 18.6)

t = 170 / 5.8

t = 29.31 s

Therefore; it will take 29.31 s for the Boxster to catch the Scion

When two point charges are a distance d part, the electric force that each one feels from the other has magnitude F. In order to make this force twice as strong, the distance would have to be changed toA) √2d
B) d/√2
C) d/4
D) 2d
E) d/2

Answers

Answer:b

Explanation:

Given

Force of attraction is F when charges are d distance apart.

Electrostatic force is given by

$F=(kq_1q_2)/(d^2)---1$

where k=constant

$q_1$ and $q_2$ are charges

d=distance between them

In order to double the force i.e. 2F

$2F=(kq_1q_2)/(d'^2)----2$

divide 1 and 2 we get

$(F)/(2F)=(d'^2)/(d^2)$

$d'=(d)/(√(2))$

Imagine that you drop an object of 10 kg, how much will be the acceleration andhow much force causes the acceleration?

Answers

If you do this on Earth, then the acceleration of the falling object is 9.8 m/s^2 ... NO MATTER what it's mass is.

If its mass is 10 kg, then the force pulling it down is 98.1 Newtons. Most people call that the object's "weight".

Singing that is off-pitch by more than about 1% sounds bad. How fast would a singer have to be moving relative to the rest of a band to make this much of a change in pitch due to the Doppler effect

Answers

Answer:

-3.396 m/s or 3.465 m/s

Explanation:

v = Speed of sound in air = 343 m/s

$v_s$ = Relative speed of the singer

f = Observed frequency

f' = Actual frequency

1% change can mean $f=1.01f'$

From the Doppler effect equation we have

$f=f'(v)/(v+v_s)\n\Rightarrow 1.01f'=f'(v)/(v+v_s)\n\Rightarrow 1.01=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(1.01)-343\n\Rightarrow v_s=-3.396\ m/s$

The velocity is -3.396 m/s

when $f=0.99f'$

$f=f'(v)/(v+v_s)\n\Rightarrow 0.99f'=f'(v)/(v+v_s)\n\Rightarrow 0.99=(343)/(343+v_s)\n\Rightarrow v_s=(343)/(0.99)-343\n\Rightarrow v_s=3.46464646465\ m/s$

The velocity is 3.465 m/s

Two ice skaters, Lilly and John, face each other while at rest, and then push against each other's hands. The mass of John is twice that of Lilly. How do their speeds compare after they push off? Lilly's speed is one-fourth of John's speed. Lilly's speed is the same as John's speed. Lilly's speed is two times John's speed. Lilly's speed is four times John's speed. Lilly's speed is one-half of John's speed.