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3 years ago

When a 600 g mass is suspended from a spring, the spring stretches 1.2 cm. What is the spring constant of the spring?

Physics

1 answer:

raketka [301]3 years ago

7 0

Well first you need to know the formula for the spring force. It’s -1/2k*d

K being the constant

D being the displacement

We can actually find the force on the spring by calculating the weight of the block. It would be mass multiplied by gravity, convert the grams to kilograms first tho.

600/1000 = 0.6kg

0.6kg * 9.8 m/s^2 = 5.88N

Now plug everything into the equation, also convert the centimeters to meters:

1.2/100 = 0.012 meters

5.88 = -1/2k*(0.012)

5.88 = k*(0.006)

We could drop the negative sign because we really just want the magnitude of the spring constant.

K = 980N

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El peso normal de un estudiante de secundaria es 725 N y el área de los dos zapatos que usa es de 412 cm2 . La presión medida qu

kotykmax [81]

Answer:

Presión = 175,97 N/m²

Explanation:

Dados los siguientes datos;

Peso del alumno (fuerza) = 725N

Área de zapatos = 412 cm² a metros cuadrados = 412/100 = 4.12 metros

Para encontrar la presión, usaríamos la siguiente fórmula;

Presión = fuerza / área

Presión = 725 / 4.12

Presión = 175,97 N/m²

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3 years ago

Which equation correctly relates mechanical energy, thermal energy, and total

Anna35 [415]

Answer:

This is because frictional energy is similar to thermal energy in a system. Therefore the thermal energy produced is equal to

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3 years ago

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

4 years ago

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more t

olga2289 [7]

The force result in stretching the spring 10.0 centimeters is 2.5N.

<h3>What is Hooke's law?</h3>

If a spring is stretched from its equilibrium position, then a force with magnitude proportional to the increase in length from the equilibrium length is pulling each end.

F = kx

where k is the proportionality constant called the spring constant or force constant.

Up to a point, the elongation of a spring is directly proportional to the force applied to it. Once you extend the spring more than 10.0 centimeters, however, it no longer follows that simple linear rule.

Let the spring constant be very low 0.04N/m

The force applied is

F = 10 cm / 0.04

F = 0.1 m / 0.04

F = 2.5 N

Thus, the force result in stretching the spring 10cm is 2.5 N.

Learn more about hooke's law.

brainly.com/question/13348278

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2 years ago

A 10-cm-thick aluminum plate (α = 97.1 × 10−6 m2/s) is being heated in liquid with temperature of 550°C. The aluminum plate has

Zarrin [17]

Answer:

356°C.

Explanation:

(1). The first step to the solution to this particular Question/problem is to determine the Biot number, and after that to check the equivalent value of the Biot number with plate constants.

That is, Biot number = (length × ∞)÷ thermal conductivity. Which gives us the answer as ∞. Therefore, the equivalent value of the ∞ on the plates constant = 1.2732 for A and 1.5708 for λ.

(2). The next thing to do is to determine the fourier number.

fourier number = [α = 97.1 × 10−6 m2/s × 15 s] ÷ (.05m)^2 = 0.5826.

(3). The next thing is to determine the temperature at the center plane after 15 s of heating.

The temperature at the center plane after 15 s of heating = 500°C [ 25°C - 500°C ] [1.2732] × e^(-1.5708)^2 ( 0.5826).

The temperature at the center plane after 15 s of heating = 356°C.

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3 years ago

When a 600 g mass is suspended from a spring, the spring stretches 1.2 cm. What is the spring constant of the spring? ​

When a 600 g mass is suspended from a spring, the spring stretches 1.2 cm. What is the spring constant of the spring?