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lions [1.4K]
2 years ago
8

An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period o

f 0.640 s. What is the spring constant of the spring?
A) 24.1 N/m
B) 2.45 N/m
C) 0.102 N/m
D) 12.1 N/m
E) 0.610 N/m
Physics
1 answer:
Viktor [21]2 years ago
6 0

Answer:

Spring constant, k = 24.1 N/m

Explanation:

Given that,

Weight of the object, W = 2.45 N

Time period of oscillation of simple harmonic motion, T = 0.64 s

To find,

Spring constant of the spring.

Solution,

In case of simple harmonic motion, the time period of oscillation is given by :

T=2\pi\sqrt{\dfrac{m}{k}}

m is the mass of object

m=\dfrac{W}{g}

m=\dfrac{2.45}{9.8}

m = 0.25 kg

k=\dfrac{4\pi^2m}{T^2}

k=\dfrac{4\pi^2\times 0.25}{(0.64)^2}

k = 24.09 N/m

or

k = 24.11 N/m

So, the spring constant of the spring is 24.1 N/m.

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Answer:

Force of 37.8 × 10^(6) N attracts the two charges

Explanation:

The force between two charges is given by

F = k*q1*q2/r²

Where q1 and q2 are 0.06 C and 0.07 C.

r is the distance between q1 and q2 which is equal to 3 m

k is a constant = 9 × 10^(9) N.m²/C²

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

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3 years ago
A freight train has a mass of [02] kg. The wheels of the locomotive push back on the tracks with a constant net force of 7.50 ×
otez555 [7]

Answer:

t = 300.3 seconds

Explanation:

Given that,

The mass of a freight train, m=1.01\times 10^7\ kg

Force applied on the tracks, F=7.5\times 10^5\ N

Initial speed, u = 0

Final speed, v = 80 km/h = 22.3 m/s

We need to find the time taken by it to increase the speed of the train from rest.

The force acting on it is given by :

F = ma

or

F=\dfrac{m(v-u)}{t}\\\\t=\dfrac{m(v-u)}{F}\\\\t=\dfrac{1.01\times 10^7\times (22.3-0)}{7.5\times 10^5}\\\\t=300.3\ s

So, the required time is 300.3 seconds.

4 0
3 years ago
A kite is 100m above the ground. if there are 200m of string out what is the angle between the string and the horizontal? (assum
jeka57 [31]

Answer :  The angle between the string and the horizontal is 30 degrees

Explanation:  Imagine this a a triangle where the length of the string (200m) is the hypotenuse and the height of the kite is the opposite side (100m) .

Let the angle between the string and the horizontal be theta.

Now  sin (Theta) = opposite side/hypotenuse

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Therefore Theta = Sin ⁻¹ ( 1/2 )

Theta = 30 degrees

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3 years ago
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3 years ago
An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. The acceleration
Debora [2.8K]

Answer:

The radius of the loop is  20.66 km

Explanation:

let the radius of the loop be r

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At the top, the pilot experiences two radial forces, which are

1) Gravitational force is  mg

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When the pilot experiences no weight,

then, mg = mv²/r

r = v² / g

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 = 20.66 x 10³3

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3 0
3 years ago
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