What is heat that is transferred by movement of a fluid

The strongest ernergy wave in the EM Spectrum is 'microwave'<br> true<br> false

Ratling [72]

Pretty sure it’s false....................

6 0

3 years ago

A 8.00-kg object is hung from the bottom end of a vertical spring fastened to an overhead beam. The object is set into vertical

Alex_Xolod [135]

Answer:

109.32 N/m

Explanation:

Given that

Mass of the hung object, m = 8 kg

Period of oscillation of object, T = 1.7 s

Force constant, k = ?

Recall that the period of oscillation of a Simple Harmonic Motion is given as

T = 2π √(m/k), where

T = period of oscillation

m = mass of object and

k = force constant if the spring

Since we are looking for the force constant, if we make "k" the subject of the formula, we have

k = 4π²m / T², now we go ahead to substitute our given values from the question

k = (4 * π² * 8) / 1.7²

k = 315.91 / 2.89

k = 109.32 N/m

Therefore, the force constant of the spring is 109.32 N/m

8 0

3 years ago

A 75 W lightbulb is being run on a 110 V outlet. Determine the resistance of the lightbulb. Provide a detailed description of th

Sauron [17]

Power is defined as

P = I*V

where I is the current and V is the voltage

Ohm's law gives us the relation betwen Voltage and current in a resistive component

V = I*R , Then

P = V² / R

We solve for R,

R = (110 V)²/ 75W = 161.33 ohms

8 0

3 years ago

Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin

tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of $m_2$ would be four times that of $m_1$ .

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is $T$ , then the angular velocity of the SHM would be

$\displaystyle \omega = \frac{2\pi}{T}$ .

Assume that the mass starts with a zero displacement and a positive velocity. If $A$ represent the amplitude of the SHM, then the displacement of the mass at time $t$ would be:

$\mathbf{x}(t) = A\sin(\omega\cdot t)$ .

The velocity of the mass at time $t$ would be:

$\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t)$ .

The acceleration of the mass at time $t$ would be:

$\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t)$ .

Let $m$ represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time $t$ would be:

$\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t)$ ,

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant $k$ will be equal to:

$\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}$ .

Since $\displaystyle \omega = \frac{2\pi}{T}$ , it can be concluded that:

$\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}$ .

For the first mass $m_1$ , if the time period is $T_1$ , then the spring constant would be:

$\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2$ .

Similarly, for the second mass $m_2$ , if the time period is $T_2$ , then the spring constant would be:

$\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Since the two springs are the same, the two spring constants should be equal to each other. That is:

$\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2$ .

Simplify to obtain:

$\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2$ .

6 0

3 years ago

A 10.00 kg mass is moving to the right with a velocity of 14.0 m/s. A 12.0 kg mass is moving to the left with a velocity of 8.00

Basile [38]

Answer:

2 m/s

Explanation:

From the conservation of momentum, the initial momentum of the system must be equal to the final momentum of the system.

Let the 10.00 kg mass be $m_1$ and the 12.0 kg mass be $m_2$ . When they collide and stick, they have a combined mass of $m_1+m_2$ .

Momentum is given by $p=mv$ . Set up the following equation:

$\displaystyle m_1v_1+m_2v_2=(m_1+m_2)v_f$ , where $v_f$ is the desired final velocity of the masses.

Call the right direction positive. To indicate the 12.0 kg object is travelling left, its velocity should be substitute as -8.00 m/s.

Solving yields:

$10\cdot 14 + 12\cdot (-8)=(10+12)v_f\\\implies v_f=\boxed{2 \text{ m/s}}$

4 0

2 years ago