PHYSICS COLLEGE

A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3

Answers

Answer 1

Answer:

Answer:

Explanation:

The formula to calculate the power is:

$P=(W)/(T)$

where

W is the work done

T is the time required for the work to be done

In the second part of the problem, we have

Work done: 3W

Time interval: 5T

So the power required is

$P=(3W)/(5T)=(3)/(5)(W)/(T)=(3)/(5)P$

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A Porsche sports car can accelerate at 8.8 m/s^2. Determine its acceleration in km/h^2.

Answers

Answer:

The acceleration expressed in the new units is $114.048 Km/h^(2)$

Explanation:

To convert from $m/s^(2)$ to $Km/h^(2)$ it is necessary to remember that there are 1000 meters in 1 kilometer and 3600 seconds in 1 hour:

Then by means of a rule of three it is get:

$8.8(m)/(s^(2)).((1Km)/(1000m)).((3600s)/(1h))^(2)$

$8.8(m)/(s^(2)).((1Km)/(1000m)).((12960000s^(2))/(1h^(2)))$

Hence, the units of meters and seconds will cancel. Notice the importance of square the ratio 3600s/1h, so that way they can match with the other units:

$114.048 Km/h^2$

So the acceleration expressed in the new units is $114.048 Km/h^2$ .

The speed of light is 3 x 10 m/s.Calculate the frequency of light that is absorbed the most by the 100m length of fibre.
Give your answer in standard form.

Answers

Answer:

$3 * 10 {}^6$

Resonances of the ear canal lead to increased sensitivity of hearing, as we’ve seen. Dogs have a much longer ear canal—5.2 cm—than humans. What are the two lowest frequencies at which dogs have an increase in sensitivity? The speed of sound in the warm air of the ear is 350 m/s.A. 1700 Hz, 3400 Hz
B. 1700 Hz, 5100 Hz
C. 3400 Hz, 6800 Hz
D. 3400 Hz, 10,200 Hz

Answers

Answer:

B. 1700 Hz, 5100 Hz

Explanation:

Parameters given:

Length of ear canal = 5.2cm = 0.052 m

Speed of sound in warm air = 350 m/s

The ear canal is analogous to a tube that has one open end and one closed end. The frequency of standing wave modes in such a tube is given as:

f(m) = m * (v/4L)

Where m is an odd integer;

v = velocity

L = length of the tube

Hence, the two lowest frequencies at which a dog will have increased sensitivity are f(1) and f(3).

f(1) = 1 * [350/(4*0.052)]

f(1) = 1682.69 Hz

Approximately, f(1) = 1700 Hz

f(3) = 3 * [350/(4*0.052)]

f(3) = 5048 Hz

Approximately, f(3) = 5100 Hz

So to deal with the irrational belief in REBT, we must Group of answer choices

A. Consult with a friend and get their feeback

B. Dispute the beliefs by asking if these are true and examining the evidence

C. Seek mental health counseling

D. It is just too hard so let's just forget it.

Answers

Answer:

i believe the answer is B

Explanation:

Seeking the right answer is the best thing to do

Sophia wanted to estimate the product 73 x 12 so she made an estimate of 70x 10. Would her estimate be greater or near the actual estimate?

Answers

her estimate would not be greater, so it would be near the actual estimate. the numbers 70 and 10 are smaller than the numbers 73 and 12, so it is not possible for 70 x 10 to be more than 73 x 12 is. i hope this helps!!

(25) A grinding machine is supported on an isolator that has two springs, each with stiffness of k and one viscous damper with damping constant of c=1.8 kNs/m. The floor on which the machine is mounted is subjected to a harmonic disturbance due to the operation of an unbalanced engine in the vicinity of the grinding machine. The floor oscillates with amplitude Y=3 mm and frequency of 18 Hz. Because of other design constraints, the stiffness of each spring must be greater than 3.25 MN/m. What is the minimum required stiffness of each of the two springs to limit the grinding machine’s steady-state amplitude of oscillation to at most 10 mm? Assume that the grinding machine and the wheel are a rigid body of weight 4200 N and can move in only the vertical direction (the springs deflect the same amount).

Answers

Answer:

k = 15.62 MN/m

Explanation:

Given:-

- The viscous damping constant, c = 1.8 KNs/m

- The floor oscillation magnitude, Yo = 3 mm

- The frequency of floor oscillation, f = 18 Hz.

- The combined weight of the grinding machine and the wheel, W = 4200 N

- Two springs of identical stiffness k are attached in parallel arrangement.

Constraints:-

- The stiffness k > 3.25 MN/m

- The grinding machine’s steady-state amplitude of oscillation to at most 10 mm. ( Xo ≤ 10 mm )

Find:-

What is the minimum required stiffness of each of the two springs as per the constraints given.

Solution:-

- The floor experiences some harmonic excitation due to the unbalanced engine running in the vicinity of the grinding wheel. The amplitude "Yo" and the frequency "f" of the floor excitation is given

- The floor is excited with a harmonic displacement of the form:

$y ( t ) = Y_o*sin ( w*t )$

Where,

Yo : The amplitude of excitation = 3 mm

w : The excited frequency = 2*π*f = 2*π*18 = 36π

- The harmonic excitation of the floor takes the form:

$y ( t ) = 3*sin ( 36\pi *t )$

- The equation of motion for the floor excitation of mass-spring-damper system is given as follows:

$m*(d^2x)/(dt^2) + c*(dx)/(dt) + k_e_q*x = k_e_q*y(t) + c*(dy)/(dt)\n\n(m)/(k_e_q)*(d^2x)/(dt^2) + (c)/(k_e_q)*(dx)/(dt) + x = y(t) + (c)/(k_e_q)*(dy)/(dt)$

Where,

m: The combined mass of the rigid body ( wheel + grinding wheel body) c : The viscous damping coefficient

k_eq: The equivalent spring stiffness of the system ( parallel )

x : The absolute motion of mass ( free vibration + excitation )

- We will use the following substitutions to determine the general form of the equation of motion:

$w_n = \sqrt{(k_e_q)/(m) } , \n\np = (c)/(2√(k_e_q*m) ) = (1800)/(2√(k_e_q*428.135) ) = (43.49628)/(√(k_e_q) )$

Where,

w_n: The natural frequency

p = ζ = damping ratio = c / cc , damping constant/critical constant

- The Equation of motion becomes:

$(1)/(w^2_n)*(d^2x)/(dt^2) + (2*p)/(w_n)*(dx)/(dt) + x = y(t) + (2*p)/(w_n)*(dy)/(dt)$

- The steady solution of a damped mass-spring system is assumed to be take the form of harmonic excitation of floor i.e:

$X_s_s = X_o*sin ( wt + \alpha )$

Where,

X_o : The amplitude of the steady-state vibration.

α: The phase angle ( α )

- The steady state solution is independent from system's initial conditions and only depends on the system parameters and the base excitation conditions.

- The general amplitude ( X_o ) for a damped system is given by the relation:

$X_o = Y_o*\sqrt{(1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2) }$

Where,

r = Frequency ratio = $(w)/(w_n) = \frac{36*\pi }{\sqrt{(k_e_q*g)/(W) } } = \frac{36*\pi }{\sqrt{(k_e_q)/(428.135) } } = (36*\pi*√(428.135) )/(√(k_e_q) )$

- We will use the one of the constraints given to limit the amplitude of steady state oscillation ( Xo ≤ 10 mm ):

- We will use the expression for steady state amplitude of oscillation ( Xo ) and determine a function of frequency ratio ( r ) and damping ratio ( ζ ):

$((X_o )/(Y_o))^2 \geq (1+ ( 2*p*r)^2)/(( 1 - r^2)^2 + ( 2*p*r)^2)\n\n((X_o )/(Y_o))^2 \geq (1+ ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)/(( 1 - ((36*\pi*√(428.135) )/(√(k_e_q) ))^2)^2 + ( 2*(43.49628)/(√(k_e_q) )*(36*\pi*√(428.135) )/(√(k_e_q) ))^2)\n\n$

$((X_o )/(Y_o))^2 \geq ( 1 + (41442858448.85813)/(k_e_q^2 ))/([ 1 - ((5476277.91201 )/(k_e_q) )]^2 + (41442858448.85813)/(k_e_q^2 ) )}\n\n((X_o )/(Y_o))^2 \geq ( (k_e_q^2 + 41442858448.85813)/(k^2_e_q ))/([ ((k_e_q - 5476277.91201)^2 )/(k_e_q^2) ] + (41442858448.85813)/(k_e_q^2 ) )}\n$

$((X_o )/(Y_o))^2 \geq ( k_e_q^2 + 41442858448.85813)/( (k_e_q - 5476277.91201)^2 +41442858448.85813 )}\n\n((10 )/(3))^2 \geq ( k_e_q^2 + 41442858448.85813)/( k^2_e_q -10952555.82402*k_e_q +3.00311*10^1^3 )}\n\n\n10.11111*k^2_e_q -121695064.71133*k_e_q +3.33637*10^1^4 \geq 0$

- Solve the inequality ( quadratic ):

$k1_e_q \geq 7811740.790197058 (N)/(m) \n\nk2_e_q \leq 4224034.972855095 (N)/(m)$

- The equivalent stiffness of the system is due to the parallel arrangement of the identical springs:

$k_e_q = (k^2)/(2k) = (k)/(2)$

- Therefore,

$k1 \geq 7811740.790197058*2 = 15.62 (MN)/(m) \n\nk2 \leq 4224034.972855095*2 = 8.448 (MN)/(m)$

- The minimum stiffness of spring is minimum of the two values:

k = 15.62 MN/m

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