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2 years ago

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At what altitude above the earth's surface is the gravitational acceleration 5. 6 m/s2?.

1 answer:

umka21 [38]2 years ago

8 0

a = G M / R^2 gravitational acceleration

R^2 = G M / a = 6.67E-11 * 5.98E24 / 5.6 m^2/s^2

R^2 = 6.67 * 5.98 / 5.6 E13 = 71.2E12

R = 8.44E6 m

H = (8.44 - 6.37) E6 m = 2.07E6m above the surface of the earth

(About 1/3 of the radius of the earth)

As a check 4/3^2 = 1.78

9.8 m/s^2 / 1.8 = 5.4 Close enough to 5.6

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u = 0.077

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Work done by friction is

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3 years ago

What type of galaxy is shown?

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3 years ago

Read 2 more answers

Two loudspeakers are 1.60 m apart. A person stands 3.00 m from one speaker and 3.50 m from the other. (a) What is the lowest fre

VMariaS [17]

Answer:

Explanation:

Given

Distance between two loud speakers $d=1.6\ m$

Distance of person from one speaker $x_1=3\ m$

Distance of person from second speaker $x_2=3.5\ m$

Path difference between the waves is given by

$x_2-x_1=(2m+1)\cdot \frac{\lambda }{2}$

for destructive interference m=0 I.e.

$x_2-x_1=\frac{\lambda }{2}$

$3.5-3=\frac{\lambda }{2}$

$\lambda =0.5\times 2$

$\lambda =1\ m$

frequency is given by

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

where $v=velocity\ of\ sound\ (v=343\ m/s)$

$f=\frac{343}{1}=343\ Hz$

For next frequency which will cause destructive interference is

i.e. $m=1$ and $m=2$

$3.5-3=\frac{2\cdot 1+1}{2}\cdot \lambda$

$\lambda =\frac{1}{3}\ m$

frequency corresponding to this is

$f_2=\frac{343}{\frac{1}{3}}=1029\ Hz$

for $m=2$

$3.5-3=\frac{5}{2}\cdot \lambda$

$\lambda =\frac{1}{5}\ m$

Frequency corresponding to this wavelength

$f_3=\frac{343}{\frac{1}{5}}$

$f_3=1715\ Hz$

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3 years ago

4) A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

VMariaS [17]

Answer:

(a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

Explanation:

Given that,

A football player starts at the 40-yard line, and runs to the 25-yard line in 2 seconds.

Suppose, the distance between 40 yard line and 25 yard line is 20 yard.

(a). We need to calculate their speed during that run

Using formula of speed

$v=\dfrac{d}{t}$

Where. d = distance

t = time

Put the value into the formula

$v=\dfrac{18.288}{2}$

$v=10\ m/s$ during that run

(b). We need to calculate their velocity

Using formula of speed

$v=\dfrac{\Delta d}{\Delta t}$

Put the value into the formula

$v=\dfrac{22.86-36.58}{2}$

$v=-6.86\ m/s$

Negative sign shows the direction of motion.

(c). If they kept running at that velocity for another 1.3 seconds,

We need to calculate the final position

Using formula of position

$d=vt$

Put the value into the formula

$d=6.86\times1.3$

$d=8.91\ m$

Hence, (a). Their speed during that run is 10 m/s.

(b). Their velocity is 6.86 m/s

(c). The final position is at 8.91 m.

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3 years ago

You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago