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Nady [450]
3 years ago
14

A 927 g mass is connected to a light spring of force constant 5 N/m that is free to oscillate on a horizontal, frictionless trac

k. The mass is displaced 8 cm from the equilibrium point and released from rest. 5 N/m 927 g 8 cm x = 0 x Find the period of the motion. Answer in units of s.

Physics
2 answers:
liraira [26]3 years ago
5 0

Answer:

2.7 s

Explanation:

The period of the motion of a massless loaded spring is given by

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

where m is the mass of the load and k is the force or spring constant.

Using values in the question and converting to appropriate units,

T = 2\pi\sqrt{\dfrac{927\times10^{-3}\text{ kg}}{5 \text{ N/m}}}= 2.7\text{ s}

In-s [12.5K]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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If a steady-state heat transfer rate of 3 kW is conducted through a section of insulating material 1.0 m2 in cross section and 2
kaheart [24]

Answer:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

Explanation:

For this case we can use the Fourier Law of heat conduction given by the following equation:

Q = -kA \frac{\Delta T}{\Delta x}   (1)

Where k = thermal conductivity = 0.2 W/ mK

A= 1m^2 represent the cross sectional area

Q= 3KW represent the rate of heat transfer

\Delta T is the temperature of difference that we want to find

\Delta x=2.5 cm =0.025 m represent the thickness of the material

If we solve \Delta T in absolute value from the equation (1) we got:

\Delta T =\frac{Q \Delta x}{Ak}

First we convert 3KW to W and we got:

Q= 3 KW* \frac{1000W}{1 Kw}= 3000 W

And we have everything to replace and we got:

\Delta T = \frac{3000 W *0.025 m}{1 m^2 (0.2 \frac{W}{mK})}= 375 K

So then the difference of temperature across the material would be \Delta T = 375 K

5 0
3 years ago
1. A spring extends by 10 cm when a mass of 100 g is attached to it. What is the spring constant? (Calculate your answer in N/m)
cupoosta [38]

Answer:

1) k = 10 [N/m]

2) a-) x = 0.4 [m]

b)  x = 0.075 [m]

Explanation:

To be able to solve this type of problems that include springs we must use Hooke's law, which relates the force to the deformed length of the spring and in the same way to the spring coefficient.

F = k*x

where:

F = force [N] (units of Newtons]

k = spring constant  [N/m]

x = distance = 10 [cm] = 0.1 [m]

Now, the weight is equal to the product of the mass by the gravity

W = m*g = F

where:

m = mass = 100 [g] = 0.1 [kg]

g = gravity acceleration = 10 [m/s²]

F = 0.1*10 = 1 [N]

Now clearing k

k = F/x

k = 1/0.1

k = 10 [N/m]

2)

a ) if the force is 4 [N]

clearing x

x = F/k

x = 4/10

x = 0.4 [m]

m = 75 [g] = 0.075 [kg]

W = m*g = F

F = 0.075*10 = 0.75 [N]

x = .75/10

x = 0.075 [m]

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3 years ago
Iron has a density of 7.9 g/cm3. What is the mass of a cube of iron with the length of one side equal to 55.0 mm?
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Answer:

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Explanation:

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Efficiency of the machine = 75%

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