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masya89 [10]
3 years ago
13

A 40.1 g object is attached to a horizontal spring with a spring constant of 11.9 N/m and released from rest with an amplitude o

f 24.7 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
Physics
1 answer:
Arlecino [84]3 years ago
3 0

Answer: v = 3.684 m/s

Explanation: The angular frequency (ω) of a loaded spring is given as

ω = √k/m

Where ω = angular frequency, k =spring constant = 11.9 N/m, m = mass of object = 40.1 g = 0.0401 kg.

The velocity of a simple harmonic motion is defied as

v = ω√A² - x²

Where A = Amplitude = 24.7cm = 0.247m and x = displacement.

For our question, we where asked to find velocity at half way, at half way, x = A/2

Hence at half way, x = 0.247/2 = 0.1235 m.

We need to get the value of angular frequency first.

ω = √(11.9/0.0401)

ω = √296.758

ω = 17.22 rad/s.

Then the velocity is

v = 17.22 √0.247² - 0.1235²

v = 17.22 √0.061009 - 0.01525225

v = 17.22 √0.04575675

v = 17.22 × 0.2139

v = 3.684 m/s

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Answer:

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3 years ago
A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0
3 years ago
A ranger in a national park is driving at 56 km/h when a deer jumps onto the road 65 m ahead of the vehicle. After a reaction ti
vagabundo [1.1K]

Answer:

 t = 1.58 s

Explanation:

given,

Speed of ranger, v = 56 km/h

                            v = 56 x 0.278 = 15.57 m/s

distance, d = 65 m

deceleration,a = 3 m/s²

reaction time = ?

using stopping distance formula

d = v. t + \dfrac{v^2}{2a}

t = \dfrac{d}{v} -\dfrac{v}{2a}

t is the reaction time

t = \dfrac{65}{15.57} -\dfrac{15.57}{2\times 3}

 t = 1.58 s

hence, the reaction time of the ranger is equal to 1.58 s.

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3 years ago
The focal length of a concave lens 13.3 cm.Calculate its power.​
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1 /13.3 = 0.07518 cm

Please mark it as Brainliest!

4 0
3 years ago
An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 f
____ [38]
<h2>a) Initial velocity = 83 ft/s</h2><h2>b) Object's maximum speed = 99.4 ft/s</h2><h2>c) Object's maximum displacement = 153.64 ft</h2><h2>d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e) The displacement is zero when t = 5.70 seconds</h2><h2>f) Object's maximum height = 153.64 ft</h2>

Explanation:

We have velocity

             v(t)= -32t + 83

Integrating

              s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

              46 = -16 x 0²+83 x 0+C

                 C = 46 feet

So displacement is

              s(t) = -16t²+83t+46

a) Initial velocity is

                 v(0)= -32 x 0 + 83 = 83 ft/s

       Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

             0 = -16t²+83t+46

             t = 5.70 seconds

Substituting in velocity equation

           v(t)= -32 x 5.70 + 83 = -99.4 ft/s

           Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

   That is

                 -32t + 83 = 0

                       t = 2.59 s

Substituting in displacement equation

                s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

   The displacement is zero when t = 5.70 seconds

f) Same as option d

   Object's maximum height = 153.64 ft

5 0
3 years ago
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