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3 years ago

Explain what do you understand by the term viscosity ?please answer me

1 answer:

5 0

Viscosity - a measure of a fluids resistance to move. It measures the internal friction of a liquid that is moving. large viscosity resists more motion, and vice versa. ( all this info is from princeton.edu)

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A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 362 m in 139 s. What

Jobisdone [24]

Answer:

$v_{avg} = 6.41 m/s$

Explanation:

Average velocity is defined as the ratio of total displacement of the motion and total time taken in that motion

here we know that initially the sky diver drops without opening parachute by total displacement 625 m

then she open her parachute and drop another 362 m

so first it took time t = 15 s to drop without open parachute

then it took t = 139 s to drop next displacement

so here total displacement is given as

$d = 625 m + 362 m$

total time is given as

$t = 15 s + 139 s$

so average velocity is given as

$v_{avg} = \frac{625 + 362}{15 + 139}$

$v_{avg} = 6.41 m/s$

4 0

3 years ago

With a mass of 109 kg, Baby Bird is the smallest monoplane ever

OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

m_1=61.4kg
m_2=109kg
v_1=1.84m/s
v_2=?

$\\ \sf\longmapsto ∆P=P$

$\\ \sf\longmapsto m_1v_1=m_2v_2$

$\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}$

$\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}$

$\\ \sf\longmapsto v_2=112.976/109$

$\\ \sf\longmapsto v_2\approx 1.3m/s$

4 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 spring hangs at rest from a support. If you suspend a 0.46 kg mass from the spring it elongates 0.79m.

Fudgin [204]

a. The restoring force in the spring has magnitude

F[spring] = k (0.79 m)

which counters the weight of the mass,

F[weight] = (0.46 kg) g = 4.508 N

so that by Newton's second law,

F[spring] - F[weight] = 0 ⇒ k = (4.508 N) / (0.79 m) ≈ 5.7 N/m

b. Using the same equation as before, we now have

F[weight] = (0.75 kg) g = 7.35 N

so that

(5.7 N/m) x - 7.35 N = 0 ⇒ x = (7.35 N) / (5.7 N/m) ≈ 1.3 m

4 0

3 years ago

PLEASE HELP!!! FIRST IS BRAINLIEST!!!

Verizon [17]

1. A Medium

2. One of these 4, the refractive index of the medium <span>Choose whichever are on the test.

3. </span>properties and nature of medium, Hope this helps.

6 0

4 years ago

Read 2 more answers

Explain what do you understand by the term viscosity ?please answer me​

Explain what do you understand by the term viscosity ?please answer me