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Ksenya-84 [330]
3 years ago
15

You are performing a knee extension exercise. You hold a 20kg weight at full knee extension. The weight is 0.4m from your knee j

oint. What is the moment created by the weight about your knee? Ignore the mass of the leg. Include the units and the direction of the moment (i.e., positive for counterclockwise, negative for clockwise).

Physics
1 answer:
dmitriy555 [2]3 years ago
7 0

Answer:

The moment is -78.4 N-m (clockwise).

Explanation:

Given:

Mass of the object (m) = 20 kg

Distance of the object from the knee joint (d) = 0.4 m

Weight of leg is not considered.

Acceleration due to gravity (g) = 9.8 m/s²

Now, weight of the object is equal to the product of its mass and acceleration due to gravity. So,

Weight = Mass × Acceleration due to gravity

            = mg=20\times 9.8 =196\ N

We know that, moment of a force about a point is defined as the product of force applied and the perpendicular distance between the point and the line of application of force.

Moment of the given weight about the knee joint is given as:

Moment about knee joint = Weight × Distance from knee joint to weight

Moment about knee joint = 196 × 0.4 = 78.4 Nm

Now, from the diagram below, we can observe that, the weight acts vertically down and thus the sense of rotation about the knee joint at point O is clockwise. So, moment is negative.

Therefore, the moment is -78.4 N-m (clockwise).

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The Event Horizon Telescope needs a 22 micro-arcsecond resolution to view the event horizon regions around black holes. If the a
likoan [24]

Answer:

14869817.395 m

Explanation:

\theta=22 microarcsecond

λ = Wavelength = 1.3 mm

Converting to radians we get

22\times 10^{-6}\frac{\pi}{180\times 3600}\ radians

From Rayleigh Criterion

\theta=1.22\frac{\lambda}{D}\\\Rightarrow D=1.22\frac{\lambda}{\theta}\\\Rightarrow D=1.22\frac{1.3\times 10^{-3}}{22\times 10^{-6}\frac{\pi}{180\times 3600}}\\\Rightarrow D=14869817.395\ m

Diameter of the effective primary objective is 14869817.395 m

It is not possible to build one telescope with a diameter of 14869817.395 m. But, we need this type of telescope. So, astronomers use an array of radio telescopes to achieve a virtual diameter in order to observe objects that are the size of supermassive black hole's event horizon.

7 0
3 years ago
When is the magnitude of the acceleration of a mass on a spring at its maximum value?
Andreas93 [3]

Answer:

A. when the mass has a speed of zero

Explanation:

In mass-spring system, the velocity and the acceleration are in anti-phase, which means that when one of the two quantities is maximum, the other one is zero, and vice-versa.

In fact:

- When the displacement of the spring is zero (x=0), the velocity is maximum, due to conservation of energy. In fact, as the displacement is zero, the elastic potential energy of the system (given by \frac{1}{2}kx^2) is zero, therefore the kinetic energy (given by \frac{1}{2}mv^2) must be maximum, and so the velocity (v) is also maximum. On the cotnrary, acceleration (a) is directly proportional to the restoring force of the spring, given by

F=-kx

so we see that when x=0, then the force is zero: F=0, and so the acceleration is zero as well.

- When the displacement of the spring is maximum, the velocity is zero, due to conservation of energy. In fact, as the displacement is maximum, the elastic potential energy of the system (given by \frac{1}{2}kx^2) is maximum, therefore the kinetic energy (given by \frac{1}{2}mv^2) must be zero, and so the velocity (v) is also zero. On the cotnrary, since acceleration (a) is directly proportional to the restoring force of the spring, given by

F=-kx

so we see that when x=maximum, then the force is maximum, and so the acceleration is maximum as well.

Based on this, the correct answer is

A. when the mass has a speed of zero


5 0
2 years ago
Read 2 more answers
the threshold of hearing is defined as the minimum discernible intensity of the sound. it is approximately 10−12w/m2 . find the
vivado [14]

The distance is 97720.5 m

From the question, we have

P = 0.06 W × 2 = 0.12 W

d = ?

Sound intensity, I = P/4πd²

I = 10⁻¹² W/m²

10⁻¹² = 0.12/4πd²

d = 97720.5 m

The distance is 97720.5 m

Sound intensity :

The power carried by sound waves per unit area in the direction  perpendicular to that region is known as sound intensity or acoustic intensity. The watt per square meter (W/m2) is the SI unit of intensity, which also covers sound intensity. Sound intensity is a measure of how quickly energy moves across a given space. The unit area in the SI measurement system is 1 m2. So Watts per square meter are used to measure sound intensity. As there will be energy flow in certain directions but not in others, sound intensity also provides a measure of direction.

To learn more about Sound intensity visit: brainly.com/question/12899113

#SPJ4

3 0
1 year ago
A 94 g particle undergoes SHM with an amplitude of 8.3 mm, a maximum acceleration of magnitude 7.8 x 103 m/s2, and an unknown ph
Lelechka [254]

Answer:

a) T = 6.49*10^-3 s

b) v = 8 m/s

c) E = 3 J

d) F = 733 N

e) F = 366.5 J

Explanation:

Given

Mass of particle, m = 94 g = 0.094 kg

Amplitude of the particle, A = 8.3 mm = 8.3*10^-3 m

Maximum acceleration of particle, a = 7.8*10^3 m/s²

the equation describing Simple Harmonic Motion is given as

x = A cos (wt +φ)

To fond the acceleration of this relationship, we would have to integrate. Twice, the first would be a Velocity, and the second acceleration that we need.

Velocity = dx/dt = -Aw sin(wt + φ)

Acceleration = d²x/dt = -Aw² cos(wt + φ)

From the question, we were given, magnitude of acceleration to be 7.8*10^3 m/s²

Aw² = 7.8*10^3

w² = 7.8*10^3 / A

w² = 7.8*10^3 / 8.3*10^-3

w² = 939759

w = √939759

w = 969

Recall, T = 2π/w, so that

T = (2 * 3.142) / 969

T = 6.49*10^-3 s

Maximum speed = Aw

Maximum speed = 8.3*10^-3 * 969

Maximum speed = 8.0 m/s

Total mechanical energy oscillator =

mgx + 1/2mx² =

1/2mv(max)² =

1/2 * 0.094 * 8² =

3 J

Maximum displacement

x = A cos(wt + φ)

For x to be maximum here, then cos(wt + φ) Must be equal to 1

Acceleration = d²x/dt² = -Aw²

And force = mass * acceleration

Force = 0.094 * 7.8*10^3

Force = 733 N

x = A cos(wt + φ), where cos(wt + φ) = 1/2

d²x/dt² = -Aw² * 1/2

d²x/dt² = 733 * 0.5

= 366.5 N

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