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3 years ago

Is it daylight all over the world at the same time then nighttime all over the world at the same time?explain.

1 answer:

5 0

It is neither. on one half of the world at a time it is daytime(because that is the side facing the sun, therefore reciving light from the sun) and on the other side on earth it would be nighttime because that side is facing away from the sun. please review this on google if you are unsure.

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If you were on a decision making board with the task of choosing which innovation to fund, what criteria would you use to make y

faust18 [17]

Explanation:

The criteria for decision making would be

1. I would fund for the school of young diabetics, for the sole purpose of them leaning and being motivated for a healthy lifestyle.

2. I would also fund for new and improved insulin pumps as old ones cause multiple problems.

3 0

3 years ago

A student wants to demonstrate entropy using the songs on her portable music player. What should she do to demonstrate the highe

frosja888 [35]

C play all the songs on shuffle. entropy has to do with randomness

3 0

2 years ago

Read 2 more answers

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

An air-track glider with a mass of 239 g is moving at 0.81 m/s on a 2.4 m long air track. It collides elastically with a 513 g g

HACTEHA [7]

Answer:

Glider it stops just when it reaches the end of the runway

Explanation:

This is a shock between two bodies, so we must use the equations of conservation of the amount of movement, in the instant before the crash and the subsequent instant, with this we calculate the second glider speed, as the shock that elastic is also keep it kinetic energy

Po = pf

Ko = Kf

Before crash

Po = m1 Vo1 + 0

Ko = ½ m1 Vo1²

After the crash

Pf = m1 Vif + Vvf

Kf = ½ m1 V1f² + ½ m2 V2f²

m1 V1o = m1 V1f + m2 V2f (1)

m1 V1o² = m1 V1f² + m2 V2f² (2)

We see that we have two equations with two unknowns, so the system is solvable, we substitute in 1 and 2

m1 (V1o -V1f) = m2 V2f (3)

m1 (V1o² - V1f²) = m2 V2f²

Let's use the relationship (a + b) (a-b) = a² -b²

m1 (V1o + V1f) (V1o -V1f) = m2 V2f²

We divide with 3 and simplify

(V1o + V1f) = V2f (4)

Substitute in 3, group and clear

m1 (V1o - V1f) = m2 (V1o + V1f)

m1 V1o - m2 V1o = m2 V1f + m1 V1f

V1f (m1 -m2) = V1o (m1 + m2)

V1f = V1o (m1-m2 / m1+m2)

We substitute in (4) and group

V2f = V1o + (m1-m2 / m1 + m2) V1o

V2f = V1o [1+ + (m1-m2 / m1 + m2)]

V2f = V1o (2m1 / (m1+m2)

We calculate with the given values

V1f = 0.81 (239-513 / 239 + 513)

V1f = 0.81 (-274/752)

V1f = - 0.295 m/s

The negative sign indicates that the planned one moves in the opposite direction to the initial one

V2f = 0.81 [2 239 / (239 + 513)]

V2f = 0.81 [0.636]

V2f = 0.515 m / s

Now we analyze in the second glider movement only, we calculate the energy and since there is no friction,

Eo = Ef

Where Eo is the mechanical energy at the lowest point and Ef is the mechanical energy at the highest point

Eo = K = ½ m2 vf2²

Ef = U = m2 g Y

½ m2 v2f² = m2 g Y

Y = V2f² / 2g

Y = 0.515²/2 9.8

Y = 0.0147 m

At this height the planned stops, let's use trigonometry to find the height at the end of the track of the track

tan θ = Y / x

Y = x tan θ

The crash occurs in the middle of the track whereby x = 1.2 m

Y = 1.2 tan 0.7

Y = 0.147 m

As the two quantities are equal in glider it stops just when it reaches the end of the runway

7 0

3 years ago

An airplane of mass 39,043.01 flies horizontally at an altitude of 9.2 km with a constant speed of 335 m/s relative to Earth. Wh

Likurg_2 [28]

Answer:

1.2 x 10¹¹ kgm²/s

Explanation:

m = mass of the airplane = 39043.01

r = altitude of the airplane = 9.2 km = 9.2 x 1000 m = 9200 m

v = speed of airplane = 335 m/s

L = Angular momentum of airplane

Angular momentum of airplane is given as

L = m v r

Inserting the values

L = (39043.01 ) (335) (9200)

L = (39043.01 ) (3082000)

L = 1.2 x 10¹¹ kgm²/s

8 0

2 years ago