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3 years ago

Derrick crawls 4 feet south, and then turns east and crawls 6 feet.

Physics

1 answer:

Anna35 [415]3 years ago

7 0

Answer:

4 is initial distance and 6 is final distance or displacement

Explanation:

displacement is the shortest distance between final and initial distance...

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If a large housefly 2.4 m away from you makes a noise of 47.0 dB, what is the noise level (in dB) of 5400 flies at that distance

dem82 [27]

Answer:

the noise level of 5400 flies is equal to 84.32 dB

Explanation:

Noise made by house flies = 47 dB from the distance of 2.4 m

to calculate the noise of 5400 flies at a distance of 2.4 m.

the intensity of noise

$10 log_{10}(\dfrac{I}{I_0}) = 47 dB$

now, I' = 5400 I₀

= $10 log_{10}(\dfrac{5400 I}{I_0})$

= $10 [log_{10}[5400] +log_{10}(\dfrac{I}{I_0}))$

= 10 ( 3.732 + 4.7 )

= 84.32 dB

hence, the noise level of 5400 flies is equal to 84.32 dB

4 0

3 years ago

How does absorption spectra occur?

Arlecino [84]

An absorption spectrum occurs when light passes through a cold, dilute gas and atoms in the gas absorb at characteristic frequencies; since the re-emitted light is unlikely to be emitted in the same direction as the absorbed photon, this gives rise to dark lines (absence of light) in the spectrum.

hope this helped :)

mark brainliest

3 0

4 years ago

A tortoise and hare start from rest and have a race. As the race begins, both accelerate forward. The hare accelerates uniformly

never [62]

Answer:

1) Vf = 3.36 m/s

2) v = 6.86 m/s

3) s = 92.3 m

4) a = - 0.23 m/s²

Explanation:

We use first equation of motion in this case:

Vf = Vi + at

where,

Vf = Final velocity = ?

Vi = Initial velocity = 0 m/s

a = acceleration = 1.4 m/s²

t = time = 2.4 s

Therefore,

Vf = 0 m/s + (1.4 m/s²)(2.4 s)

Vf = 3.36 m/s

We again use first equation of motion but with t= 4.9 s now:

Vf = 0 m/s + (1.4 m/s²)(4.9 s)

Vf = 6.86 m/s

Now, this velocity remained constant for net 11 seconds. Hence, the velocity of hare after 8.9 s is:

v = 6.86 m/s

First we use second equation of motion to find distance covered in accelerated motion:

s₁ = Vi t + (0.5)at²

s₁ = (0 m/s)(4.9 s) + (0.5)(1.4 m/s²)(4.9 s)²

s₁ = 16.8 m

Now, we calculate the distance covered in uniform motion:

s₂ = vt

s₂ = (6.86 m/s)(11 s)

s₂ = 75.5 m

Now, total distance covered before slowing down is given as:

s = s₁ + s₂ = 16.8 m + 75.5 m

s = 92.3 m

Using third equation of motion for the decelerated motion:

2as = Vf² - Vi²

where,

a = deceleration = ?

s = distance covered = 102 m

Vf = Final Speed = 0 m/s

Vi = Initial Speed = 6.86 m/s

Therefore,

(2)(a)(102 m) = (0 m/s)² - (6.86 m/s)²

a = - (47.06 m²/s²)/(204 m)

a = - 0.23 m/s²

3 0

4 years ago

Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble C

Serhud [2]

Answer:

The current value of the Hubble's constant = 73 km/sec/Mpc.

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Explanation:

The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.

Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.

Hence,

v is directly proportional to d

where, v = apparent velocity

d = distance

if we equate velocity and distance then there comes Hubble constant.

v = $H_{0}$ x d

$H_{0}$ = 73 km/sec/Mpc

where, Mpc = Mega Parsec = 1 Mpc = 3.086 x $10^{19}$ km

We can use Hubble constant to tell the age of universe.

t = d/v

t = d/( $H_{0}$ xd)

t = 1/ $H_{0}$

Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.

Now, if we hypothetically change the value of Hubble constant,

from $H_{0}$ = 73 km/sec/Mpc to $H_{0}$ = 700 km/sec/Mpc

then the age of universe will be:

t = 1/ $H_{0}$

first convert the units of new $H_{0}$ into 1/s

$H_{0}$ = (700) x (/3.08 x $10^{19}$ )

$H_{0}$ = 227.27 x $10^{-19}$ = 2.27 x $10^{-21}$ 1/s

So,

Age of universe will be:

t = 1/ $H_{0}$ = 1/2.27x $10^{-21}$ 1/s

t = 2.27 x $10^{21}$ s

t = 71.9 trillion years

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

6 0

3 years ago

A wave on a string has a wavelength of 0.90 m at a frequency of 600 Hz. If a new wave at a frequency of 300 Hz is established in

kobusy [5.1K]

Answer:

λ₂ = 1.8 m

Explanation:

given,

wavelength of the string 1 = 0.90 m

frequency of the string 1 = 600 Hz

wavelength of string 2 = ?

frequency of the string 2 = 300 Hz

we now,

$f\ \alpha\ \dfrac{1}{\lambda}$

now,

$\dfrac{f_1}{f_2}=\dfrac{\lambda_2}{\lambda_1}$

$\dfrac{600}{300}=\dfrac{\lambda_2}{0.9}$

λ₂ = 2 x 0.9

λ₂ = 1.8 m

Hence, the wavelength of the second string is equal to λ₂ = 1.8 m

8 0

3 years ago