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2 years ago

13

What does gravitational potential energy typically change into? Question 8 options: A-Electric energy B- Kinetic energy C- Radio

active energy D-Radioactive energy

2 answers:

MariettaO [177]2 years ago

7 0

The answer is “B”, Kinetic Energy

nika2105 [10]2 years ago

6 0

<em>Answer:</em>

<em>kinetic energy </em>

<em>Explanation:</em>

<em>i should know i had a test about it.</em>

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What is the name of the database that Tim Berners Lee built??

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single goose sounds a loud warning when an intruder enters the farmyard. Some distance from the goose, you measure the sound lev

Goryan [66]

Answer:

The sound level of the 26 geese is $Z_{26}= 96.15 dB$

Explanation:

From the question we are told that

The sound level is $Z_1 = 81.0 \ dB$

The number of geese is $N = 26$

Generally the intensity level of sound is mathematically represented as

The intensity of sound level in dB for one goose is mathematically represented as

$Z_1 = 10 log [\frac{I}{I_O} ]$

Where I_o is the threshold level of intensity with value $I_o = 1*10^{-12} \ W/m^2$

$I$ is the intensity for one goose in $W/m^2$

For 26 geese the intensity would be

$I_{26} = 26 * I$

Then the intensity of 26 geese in dB is

$Z_{26} = 10 log[\frac{26 I }{I_o} ]$

$Z_{26} = 10 log (\ \ 26 * [\frac{ I }{I_o} ]\ \ )$

$Z_{26} = 10 log (\ \ 26 \ \ ) * (\ \ 10 log [\frac{ I }{I_o} ]\ \ )$

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$Z_{26} = 10 log 26 + 10 log [\frac{I}{I_0} ]$

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2 years ago

A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250

Leviafan [203]

Answer:x=23.4 cm

Explanation:

Given

mass of block $m=0.5 kg$

inclination $\theta =30$

coefficient of static friction $\mu =0.35$

coefficient of kinetic friction $\mu _k=0.25$

distance traveled $d=77.3 cm$

spring constant $k=35 N/m$

work done by gravity+work done by friction=Energy stored in Spring

$mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}$

$mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}$

$0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}$

$x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}$

$x=0.234 m$

$x=23.4 cm$

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2 years ago