Find f (the force) in relation to moments.

Please help me with this

Arturiano [62]

I can’t see the full pic

5 0

3 years ago

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What is the passenger's apparent weight at t=1.0s?

Fittoniya [83]

Answer:

For the complete question provided in explanation, if the elevator moves upward, then the apparent weight will be 1035 N. While for downward motion the apparent weight will be 435 N.

Explanation:

The question is incomplete. The complete question contains a velocity graph provided in the attachment. This is the velocity graph for an elevator having a passenger of 75 kg.

From the slope of graph it is clear that acceleration at t = 1 sec is given as:

Acceleration = a = (4-0)m/s / (1-0)s = 4 m/s^2

Now, there are two cases:

1- Elevator moving up

2- Elevator moving down

For upward motion:

Apparent Weight = m(g + a)

Apparent Weight = (75 kg)(9.8 + 4)m/s^2

<u>Apparent Weight = 1035 N</u>

For downward motion:

Apparent Weight = m(g - a)

Apparent Weight = (75 kg)(9.8 - 4)m/s^2

<u>Apparent Weight = 435 N</u>

4 0

3 years ago

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A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10

IRINA_888 [86]

Complete Question:

A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to

Answer:

t= 16.7 sec.

Explanation:

As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:

γ = (ωf -ω₀) / t

If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:

γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².

When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:

γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec

8 0

3 years ago

Very short pulses of high-intensity laser beams are used to repair detached portions of the retina of the eye. The brief pulses

Tom [10]

Answer:

a) U = 0.375 mJ

b) p_rad = 4.08 mPa

c) λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d) E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

Explanation:

Given:

λ_air : wavelength = 810 * 10^(-9) m

P: Power delivered = 0.25 W

d : Diameter of circular spot = 0.00051 m

c : speed of light vacuum = 3 * 10^8 m/s

n_air : refraction Index of light in air = 1

n_med : refraction Index of light in medium = 1.34

ε_o : permittivity of free space = 8.85 * 10^-12 C / Vm

part a

The Energy delivered to retina per pulse given that laser pulses are 1.50 ms long:

U = P*t

U = (0.25 ) * (0.0015 )

U = 0.375 mJ

Answer : U = 0.375 mJ

part b

What average pressure would the pulse of the laser beam exert at normal incidence on a surface in air if the beam is fully absorbed?

p_rad = I / c

Where I : Intensity = P / A

p_rad = P / A*c

Where A : Area of circular spot = pi*d^2 / 4

p_rad = 4P / pi*d^2*c

p_rad = 4(0.25) / pi*0.00051^2*(3.0 * 10^8)

p_rad = 0.00408 Pa

Answer : p_rad = 4.08 mPa

part c

What are the wavelength and frequency of the laser light inside the vitreous humor of the eye?

λ_med = n_air*λ_air / n_med

λ_med = (1) * (810 nm) / 1.34

λ_med = 604 nm

f_med = f_air

f_med = c / λ_air

f_med = (3*10^8) / (810 * 10^-9)

f_med = 3.7 * 10^14 Hz

Answer : λ_med = 604 nm ; f_med = 3.7 * 10^14 Hz

d)

What is the electric and magnetic field amplitude in the laser beam?

I = P / A

I = 0.5*ε_o*c*E_o ^2

I = 4P / pi*d^2

Hence, E_o = ( 8 P / ε_o*c*pi*d^2 ) ^ 0.5

E_o = ( 8 * 0.25 / (8.85*10^-12) * (3*10^8) * π * (0.00051)^2) ^ 0.5

E_o = 3.04 * 10^4 V / m

For maximum magnetic field strength:

B_o = E_o / c

B_o = 3.04 * 10^4 / (3*10^8)

B _o = 1.013 *10^-4 Nm/Amp

Answer: E_o = 3.04 * 10^4 V / m , B _o = 1.013 *10^-4 Nm/Amp

5 0

3 years ago

What happens to the forces acting on a falling object at terminal velocity?

xenn [34]

Answer: Objects falling through a fluid eventually reach terminal velocity. At terminal velocity, the object moves at a steady speed in a constant direction because the resultant force acting on it is zero.

8 0

3 years ago

Find f (the force) in relation to moments.​

Find f (the force) in relation to moments.