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3 years ago

What do velocity and acceleration have in common?

Physics

2 answers:

Allushta [10]3 years ago

8 0

Acceleration is the change in velocity

Igoryamba3 years ago

4 0

They both products of speed

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A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,

kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

$\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}$

so that the x-component of $\vec v_{\rm ave}$ is

$\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}$

and its y-component is

$\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}$

Solve for $x_2$ and $y_2$ , which are the x- and y-components of the copter's position vector after t = 1.60 s.

$x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}$

$y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}$

Note that I'm reading the given details as

$x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}$

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0

2 years ago

7. A 50 kg pole vaulter running at 10 m/s vaults over the bar. Her speed when she is over the bar is 1.0 m/s. Neglect air resist

luda_lava [24]

Answer: 5.05 m

Explanation:

4 0

2 years ago

Read 2 more answers

Define anterior and posterior in correlation to the body. (ANATOMY)

tatyana61 [14]

Answer:

it should be right it's from go.ogle hm!!!

Explanation:

Anterior or ventral - front (example, the kneecap is located on the anterior side of the leg). Posterior or dorsal - back (example, the shoulder blades are located on the posterior side of the body). Medial - toward the midline of the body (example, the middle toe is located at the medial side of the foot).

8 0

2 years ago

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The following equation is an example of decay? 232/90 TH---4/2 HE +228/88 RA?

DaniilM [7]

Answer:

Alpha decay

Explanation:

Alpha decay is one of the three major types of decays, others being, beta decay and gamma decay.
When a radioactive isotope undergoes alpha decay it emits alpha particles. An alpha particle is equivalent to the nucleus of Helium atom.
Therefore, an atom undergoing decay, its atomic mass is decreased by 4 and its atomic number is decreased by 2.
Thus, since 232/90 Th, has undergone alpha decay its mass number is reduced by 4 to 228 and its atomic number by 2 to 88, and becomes 228/88 Ra.

5 0

2 years ago

A stereo speaker produces a pure \"c\" tone, with a frequency of 262 hz. what is the period of the sound wave produced by the sp

jasenka [17]

Answer:

3.82 ms

Explanation:

The period of a wave is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f is the frequency.

In this problem, f = 262 Hz, so the period if this sound wave is

$T=\frac{1}{262 Hz}=3.82\cdot 10^{-3} s=3.82ms$

4 0

2 years ago