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4 years ago

How many meters are in 55 kilometers? Express your answer using two significant figures.

Physics

2 answers:

Shkiper50 [21]4 years ago

4 0

Answer:

nsndnsnsnddnndndndndjdjdndnndndndjdjdjdjdjdjdjjdjdjd

miss Akunina [59]4 years ago

4 0

5.5 x 10^4 ( 10 power 4)

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A GPS receiver can tell you your exact latitude and longitude?

Dmitry_Shevchenko [17]

That is false. A GPS cant tell latitude and longitude

3 0

3 years ago

Two sound waves, from two different sources with the same frequency, 540 Hz, travel in the same direction at 330 m s . The sourc

oee [108]

Answer:

The value is $\Delta \phi = 4.12 \ rad$

Explanation:

From the question we are told that

The frequency of each sound is $f_1 = f_2 = f = 540 \ Hz$

The speed of the sounds is $v = 330 \ m/s$

The distance of the first source from the point considered is $a = 4.40 \ m$

The distance of the second source from the point considered is $b = 4.00 \ m$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_a = 2 \pi [\frac{a}{\lambda} + ft]$

Generally the phase angle made by the first sound wave at the considered point is mathematically represented as

$\phi_b = 2 \pi [\frac{b}{\lambda} + ft]$

Here b is the distance o f the first wave from the considered point

Gnerally the phase diffencence is mathematically represented as

$\Delta \phi= \phi_a - \phi_b = 2 \pi [\frac{ a}{\lambda} + ft ] - 2 \pi [\frac{b}{\lambda} + ft ]$

=> $\Delta \phi = \frac{2\pi [ a - b]}{ \lambda }$

Gnerally the wavelength is mathematically represented as

$\lambda = \frac{v}{f}$

=> $\lambda = \frac{330}{540}$

=> $\lambda = 0.611 \ m$

=> $\Delta \phi = \frac{2* 3.142 [ 4.40 - 4.0 ]}{ 0.611 }$

=> $\Delta \phi = 4.12 \ rad$

5 0

3 years ago

Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured

kvv77 [185]

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

Mass of the machine, m=70 kg
Stiffness of the system, k=30000 N/m
Damping ratio=0.2
Damping force, F=450 N
Angular velocity $\omega=13\ \rm rad/s$

a)We know that the amplitude X at steady state is given by

$X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}\\$

Where

$\omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s$

$\omega=13\ \rm rad/s$
$\rho=0.2$

$F_0=450\ \rm N$
$m=70\ \rm kg$

$X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m$

b) The phase shift of the motion is given by

$\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393\\$

c)Transmissibility ratio is given by

$T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57$

d)The magnitude of the force transmitted to the ground is

$F_T=(T.R)\times F_0\\\\=450\times1.57\\\\=707.6\ \rm N$

e)The maximum velocity is given by $V_{max}$

$V_{max}=\omega A_0\\\\=13\times 0.0229\\\\=0.298\ \rm m/s$

6 0

3 years ago

The strength of the force of gravity depends on a. the masses of the objects and their speeds. b. the masses of the objects and

Lostsunrise [7]

Gravitational force = Gxm1xm2 /r^2 This equation depends only upon object's masses and distance between them. So option b is correct

5 0

3 years ago

HELP: I’ve been stuck on this problem for a while now.

Arlecino [84]

Answer:

a.work done=force *displacement

=500N*46m

=23000 Joule

b.power=work done/time taken

=23000/25

=920 watt

c.GPE=m*g*h(m=mass,g=gravity due to acceleration,h=height)

=60kg*9.8m/s*14m

=8232 joule

Explanation:

3 0

4 years ago