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WARRIOR [948]
2 years ago
8

What is the difference between synchronous and asynchronous?.

Physics
2 answers:
puteri [66]2 years ago
5 0
Synchronous communication are phone calls or video meetings. Asynchronous communication happens when information can be exchanged independent of time. It doesn't require the recipient's immediate attention, allowing them to respond to the message at their convenience.

Hope this helps!
Please give Brainliest!
serg [7]2 years ago
3 0

Answer:synchronous communications are scheduled, real-time interactions by phone, video, or in-person. Asynchronous communication happens on your own time and doesn't need scheduling.

Explanation:

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This refers to the ability of the joints to move through a full range of motion
Sati [7]

Answer:

Flexibility is the ability of a joint or series of joints to move through an unrestricted, pain free range of motion. ... These soft tissues include: muscles, ligaments, tendons, joint capsules, and skin.

Explanation:

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2 years ago
What else is produced when sodium carbonate decomposes?
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Answer: What else is produced when sodium carbonate decomposes?

Explanation:

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On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed som
k0ka [10]

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

v=f \lambda

Where:

v is the speed

f=\frac{1}{T} is the frequency, which has an inverse relation with the period T=1 h

\lambda=800 km is the wavelength

Solving with the given units:

v=\frac{1}{T}\lambda

v=\frac{1}{1 h}800 km

v=800.00 km/h This is the speed of the wave in km/h

Transforming this speed to m/s:

v=800.00 \frac{km}{h} \frac{1 h}{3600 s} \frac{1000 m}{1 km}

v=222.22 m/s This is the speed of the wave in m/s

5 0
3 years ago
In a local diner, a customer slides an empty coffee cup down the counter for a refill. The cup slides off the counter and strike
zysi [14]

a) t=\sqrt{\frac{2h}{g}}

b) v=\frac{d}{\sqrt{\frac{2h}{g}}}

c) v=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d) \theta=tan^{-1}(\frac{2h}{d}) (radians)

Explanation:

a)

The motion of the cup sliding off the counter is the motion of a projectile, consisting of two independent motions:

- A uniform motion along the horizontal direction

- A uniformly accelerated motion (free fall) along the vertical direction

The time of flight of the cup is entirely determined by the vertical motion, therefore we can use the suvat equation:

s=ut+\frac{1}{2}at^2

where here:

s=h (the vertical displacement is the height of the counter)

u=0 (the initial vertical velocity of the cup is zero)

a=g (the vertical acceleration is the acceleration of gravity)

Solving for t, we find the time of flight of the cup:

h=0+\frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}

b)

To solve this part, we just analyze the horizontal motion of the cup.

Here we know that the horizontal motion of the cup is uniform: this means that is horizontal speed is constant during the whole motion, and it is actually equal to the speed at which the mug leaves the counter.

For a uniform motion, the speed is given by

v=\frac{d}{t}

where

d is the distance covered

t is the time taken

Here, the distance covered is d, the distance from the base of the counter, while the time taken is the time of flight:

t=\sqrt{\frac{2h}{g}}

Substituting into the previous equation, we find the speed of the mug as it leaves the counter:

v=\frac{d}{\sqrt{\frac{2h}{g}}}

c)

Here we want to find the speed of the cup immediately before it hits the floor.

Here we have to consider that while the mug falls, its vertical speed increases, while the horizontal speed remains constant.

Therefore, the horizontal speed of the cup before it hits the ground is:

v_x=\frac{d}{\sqrt{\frac{2h}{g}}}=d\sqrt{\frac{g}{2h}}

The vertical speed instead is given by the suvat equation:

v_y=u_y + at

where:

u_y=0 is the initial vertical velocity

a=g is the acceleration

t=\sqrt{\frac{2h}{g}} is the time of flight

Substituting,

v_y = 0 +g(\sqrt{\frac{2h}{g}})=\sqrt{2gh}

The actual speed of the cup just before it hits the floor is the resultant of the horizontal and vertical speeds, so it is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{d^2(\frac{g}{2h})+(2gh)}

d)

Just before hitting the floor, the velocity of the cup has two components:

v_x=d\sqrt{\frac{g}{2h}} is the horizontal component (in the forward direction)

v_y=\sqrt{2gh} is the vertical component (in the downward direction)

Since the two components are perpendicular to each other, the angle of the direction is given by the equation

tan \theta = \frac{v_y}{v_x}

where here \theta is measured as below the horizontal direction.

Substituting the expressions for v_x,v_y, we find:

tan \theta = \frac{\sqrt{2gh}}{d\sqrt{\frac{g}{2h}}}=\frac{2h}{d}

So

\theta=tan^{-1}(\frac{2h}{d}) (radians)

4 0
3 years ago
How far can your little brother get if he can travel at 2.5 m/s and in 5?
KatRina [158]

d=? v=2.5 u=0 and t=5 therefore the formula to be used to find the distance my brother covered is d=1/2(v-u)t

d=1/2(2.5-0)5

=6.15m

4 0
2 years ago
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