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3 years ago

A wave has a frequency of 775 Hz. What is its period?

Physics

1 answer:

Leto [7]3 years ago

4 0

The period is simply the inverse of the frequency, therefore:

T = 1 / f

T = 1 / 775 Hz

T = 0.001290 s → possible answer

T = 1.29 × 10⁻³ s → possible answer

T = 1.29 ms → possible answer

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drek231 [11]

Answer:

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Explanation:

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2 years ago

An airplane is heading due south at a speed of 560 km/h . If a wind begins blowing from the southwest at a speed of 80.0 km/h (a

polet [3.4K]

Answer:

the magnitude of Vpg = 493.711 km/h

Explanation:

given data

speed Vpg = 560 km/h

speed Vwg = 80 km/h

solution

we get here magnitude of the plane velocity w.r.t. ground is

we know that the Vpg = Vpw + Vwg .....................1

writing the component of the velocity that is

Vpw = (0 km/h î - 560 km/h j )

Vwg = (80 cos 45 km/h î + 80 sin 45 km/h j)

adding these

Vpg = (0+80 cos 45 km/h ) î + ( -560 + 80 sin 45 km/h j)i

Vpg = (42.025 ) î (-491.92 km/h)j

now we take magnitude

the magnitude of Vpg = $\sqrt{(42.025^2+(-491.92)^2)} km/h$

the magnitude of Vpg = 493.711 km/h

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3 years ago

A ball is whirled on the end of a string in a horizontal circle of radius R at speed v. By which one of the following means can

Vera_Pavlovna [14]

Explanation:

When an object moves in a circular path, it will have circular acceleration. Its magnitude of acceleration is given by :

$a=\omega^2R$

Since, $\omega=\dfrac{2\pi }{T}R$

$a=(\dfrac{2\pi}{T})^2R$

T is the time period

R is the radius of the circular path

To increase the centripetal acceleration bu a factor of 1.5 or 3/2, radius of circle must be increase by a factor of 6 and T is increased by a factor of 2 such that,

R'=6R and T'=2T

So,

$a'=(\dfrac{2\pi}{T'})^2R'$

$a'=(\dfrac{2\pi}{(2T)})^2(6R)$

$a'=\dfrac{6}{4}(\dfrac{2\pi}{T})^2R$

$a'=\dfrac{3}{2}(\dfrac{2\pi}{T})^2R$

Hence, this is the required solution.

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3 years ago

A ball is dropped from the roof of a 25-m-tall building. What is the velocity of the object when it touches the ground? Suppose

Wewaii [24]

Answer:

a) $h=25m$

b) $v=19.8m/sec$

Explanation:

From the question we are told that:

Height $h=25m$

Bounce Height $h'=20m$

Generally the Kinematic equation is mathematically given by

$V=\sqrt{2gh}\\\\V=\sqrt{2*9.81*25}$

$V=22.1m/sec$

Therefore Height

$h=\frac{V^2}{2g}\\\\h=\frac{22.1^2}{2*9.81}$

$h=25m$

Generally the Kinematic equation is mathematically given by

$v^2=2ah$

$v^2=2*9.8*20$

$v=\sqrt{2*9.8*20}$

$v=19.8m/sec$

3 0

3 years ago