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Natalija [7]
2 years ago
14

In a supermarket, you place a 22.3-N (around 5 lb) bag of oranges on a scale, and the scale starts to oscillate at 2.7 Hz. What

is the force constant (spring constant) of the spring of the scale?
Physics
1 answer:
allsm [11]2 years ago
7 0

Answer:

Force constant, k = 653.3 N/m

Explanation:

It is given that,

Weight of the bag of oranges on a scale, W = 22.3 N

Let m is the mass of the bag of oranges,

m=\dfrac{W}{g}

m=\dfrac{22.3}{9.8}

m = 2.27 kg

Frequency of the oscillation of the scale, f = 2.7 Hz

We need to find the force constant (spring constant) of the spring of the scale. We know that the formula of the frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

k=4\pi^2 f^2m

k=4\pi^2 \times (2.7)^2\times 2.27

k = 653.3 N/m

So, the force constant of the spring of the scale is 653.3 N/m. Hence, this is the required solution.

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The solution for this problem is using the formula: m1V1=m2V2 but we need to get V1:

V1= (m2/m1) V2


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2 years ago
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Explanation:

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A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit
inn [45]

Answer:twice of initial value

Explanation:

Given

spring compresses x_1 distance for some initial speed

Suppose v is the initial speed and k be the spring constant

Applying conservation of energy

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\dfrac{1}{2}mv^2=\dfrac{1}{2}kx_1^2----1

When speed doubles

\dfrac{1}{2}m(2v)^2=\dfrac{1}{2}kx_2^2----2

divide 1 and 2

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7 0
2 years ago
A 5⁢kg object is released from rest near the surface of a planet such that its gravitational field is considered to be constant.
Umnica [9.8K]

Answer:

The gravitational force exerted on the object is 75 N (answer D)

Explanation:

Hi there!

The gravitational force is calculated as follows:

F = m · g

Where:

F = force of gravity.

m = mass of the object.

g = acceleration due to gravity (unknown).

For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:

y = y0 + v0 · t + 1/2 · a · t²

Where:

y = position at time t.

y0 = initial position.

v0 = initial velocity.

t = time.

g = acceleration due to gravity.

Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.

Then, after 2 seconds, the height of the object will be -30 m:

y = y0 + v0 · t + 1/2 · g · t²

-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²

-30 m = 1/2 · g · 4 s²

-30 m = 2 s ² · g

-30 m/2 s² = g

g = -15 m/s²

Then, the magnitude of the gravitational force will be:

F = m · g

F = 5 kg · 15 m/s²

F = 75 N

The gravitational force exerted on the object is 75 N (answer D)

Have a nice day!

8 0
2 years ago
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Check the attached file for the solution.

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