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4 years ago

12

In advertising, which aspect does music help to address when a brand uses different types of music for advertising in the United

States and in Asia?

1 answer:

Oksi-84 [34.3K]4 years ago

5 0

im pretty sure its entertainment

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An athlete performing a long jump leaves the ground at a 32.9 ∘ angle and lands 7.78 m away.part a )what was the takeoff speed ?

MatroZZZ [7]

Answer:

Explanation:

Given

Inclination $\theta =32.9^{\circ}$

Distance of landing point $R=7.78\ m$

Considering athlete to be an Projectile

range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

where u=launch velocity

$7.78=\frac{u^2\sin (2\times 32.9)}{9.8}$

$u^2=83.58$

$u=\sqrt{83.58}$

$u=9.142\ m/s$

(b)If u is increased by 8% then

new velocity is $u'=9.87\ m/s$

$R'=\frac{u'^2\sin 2\theta }{g}$

$R'=9.07\ m$

3 0

4 years ago

Read 2 more answers

A meteorite has a speed of 90.0 m/s when 850 km above the Earth. It is falling vertically (ignore air resistance) and strikes a

Schach [20]

Before it hits the sand bed, the meteorite is accelerating uniformly with $g=9.80\dfrac{\rm m}{\mathrm s^2}$ , so that its speed $v$ satisfies

$v^2-{v_0}^2=-2g\Delta y$

where $v_0=90.0\dfrac{\rm m}{\rm s}$ is its initial speed and $\Delta y=(0-850)\,\mathrm{km}=-850\,\mathrm{km}$ is its change in altitude. Notice that we're taking the meteorite's starting position in the atmosphere to be the origin, and the downward direction to be negative. Now,

$v^2-\left(-90.0\dfrac{\rm m}{\rm s}\right)^2=-2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(-850,000\,\mathrm m)\implies\boxed{v=4080\dfrac{\rm m}{\rm s}}$

3 0

3 years ago

A severe storm on January 10, 1992, caused a cargo ship near the Aleutian Islands to spill 29,000 rubber ducks and other bath to

Anna35 [415]

The distance that the rubber ducks traveled = 1600 miles.
The time taken to travel this distance = 10 months.

By definition, the approximate average speed of the ocean is
v = (1600 miles)/(10 months)

Note that
1 mile = 1609 m
10 months = (10 months)*(30 days/month) *(24 hours/day)
= 7200 hours (approx.)
= 7200*3600 = 2.592 x 10⁷ s

Therefore
$v = \frac{(1600 \, m)*(1609 \, \frac{m}{mi}) }{(10 \, months)*(2.592 \times 10^{7} \, \frac{s}{10 \, months}) } = 0.0993 \, \frac{m}{s}$

Also,
$v = \frac{1600 \, m}{7200 \, h} =0.222 \, \frac{mi}{h}$

Answer:
The average velocity is approximately 0.1 m/s, or 0.222 m/h.

6 0

3 years ago

What is the unit for height

juin [17]

The unit of height is:
Feet
Inches
Centimeters

5 0

3 years ago

Read 2 more answers

A parallel-plate capacitor stores charge Q. The capacitor is then disconnected from its voltage source, and the space between th

Stells [14]

Answer:

The relationship between the initial stored energy $PE_{i}$ and the stored energy after the dielectric is inserted $PE_{f}$ is:

c) $PE_{f} =0.5\ PE_{i}$

Explanation:

A parallel plate capacitor with $C_{o}$ that is connected to a voltage source $V_{o}$ holds a charge of $Q_{o} =C_{o} V_{o}$ . Then we disconnect the voltage source and keep the charge $Q_{o}$ constant . If we insert a dielectric of $\kappa=2$ between the plates while we keep the charge constant, we found that the potential decreases as:

$V=\frac{V_{o}}{\kappa}$

The capacitance is modified as:

$C=\frac{Q}{V} =\kappa\frac{Q_{o}}{V_{o}}=\kappa\ C_{o}$

The stored energy without the dielectric is

$PE_{i}=\frac{1}{2}\frac{Q_{o}^{2}}{C_{o}}=\frac{1}{2}C_{o}V_{o}^{2}$

The stored energy after the dielectric is inserted is:

$PE_{f}=\frac{1}{2}\frac{Q^{2}}{C}=\frac{1}{2}CV^{2}$

If we replace in the above equation the values of V and C we get that

$PE_{f}=\frac{1}{2}\kappa\ C_{o}(\frac{V_{o}}{\kappa})^{2}=\frac{1}{\kappa}(\frac{1}{2}C_{o}V_{o}^{2})$

$PE_{f} =\frac{PE_{i}}{\kappa}$

Finally

$PE_{f} =0.5\ PE_{i}$

5 0

3 years ago