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madam [21]
3 years ago
15

(10 points) A spring with a 7-kg mass and a damping constant 12 can be held stretched 1 meters beyond its natural length by a fo

rce of 4 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c2−4mk? m2kg2/sec2 Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c1eαt+c2eβt where α= (the larger of the two) β=
Physics
1 answer:
Ierofanga [76]3 years ago
8 0

Answer:

......................

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An automobile having a mass of 2000 kg deflects its suspension springs 0.02 m under static conditions. Determine the nafural fre
alexandr1967 [171]

Answer:Frequency = 3.525 Hertz

Explanation:In static equilibrium, kd =mg

Where k= effective spring constant of the spring.

mg= The weight of the car.

d= static deflection.

Therefore, w =SQRTg/d

w = SQRT 9.81/0.02

w= 22.15 rad/sec

Converting to Hertz unit for frequency

1 rad/s = 0.1591

22.15rad/s=?

22.15 × 0.1591= 3.525 hertz

7 0
3 years ago
Which will most likely result in the revision of a theory?
viva [34]
I would say A not 100℅ thou
4 0
3 years ago
Read 2 more answers
Can anyone help with these questions please
Svet_ta [14]
Here... I've finished this course already

6 0
3 years ago
The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).
sammy [17]

Answer:

(a) emf_L=-LI_{max}\omega cos(\omega t)

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

I(t)=I_{max}sin(\omega t)   (1)

(a) The induced emf is given by the following formula

emf_L=-L\frac{dI}{dt}    (2)

You derivative the expression (1) in the expression (2):

emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

emf_L=-\omega LI_{max}

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully

6 0
3 years ago
13. A step up transformer has 250 turns on its primary and 500 turns on it secondary. When the primary is connected to a 200 V a
Lisa [10]

Answer:

The output power is 2 kW

Explanation:

It is given that,

Number of turns in primary coil, N_p=250

Number of turns in secondary coil, N_s=500

Voltage of primary coil, V_p=200\ V

Current drawn from secondary coil, I_s=5\ A

We need to find the power output. It is equal to the product of voltage and current. Firstly, we will find the voltage of secondary coil as :

\dfrac{N_p}{N_s}=\dfrac{V_p}{V_s}

\dfrac{250}{500}=\dfrac{200}{V_s}

V_s=400\ V

So, the power output is :

P_s=V_s\times I_s

P_s=400\ V\times 5\ A

P_s=2000\ watts

or

P_s=2\ kW

So, the output power is 2 kW. Hence, this is the required solution.

8 0
3 years ago
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