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Ahat [919]
2 years ago
8

A particle of mas 0.1 kg moves along the trajectory x(t)-3t t where x is in meters and t is in seconds. What is the kinetic ener

gy of the particle at time t? Select one: a. 0.3 (t 1)3 o b. 0.25 (t2 -1) c. 0.25 (t + 1)2 d. 0.45 (t2+1)2 o e. 3.0 (1-)
Physics
1 answer:
devlian [24]2 years ago
6 0

Answer:

Kinetic energy = 0.45 joules

Explanation:

We have velocity of the particle is

v=\frac{dx(t)}{dt}\\\\v=\frac{d(3t)}{dt}\\v=3m/s

Now kinetic energy(K.E) is given by

K.E=\frac{1}{2}mv^{2}

applying values we get

Kinetic energy =\frac{1}{2}0.1\times 3^{2}\\\\\therefore K.E=0.45Joules

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if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 m, how much sound power does that person generate
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The sound power the person generated is 2.313 \times 10^{-5} \ W.

<h3>Area of the sound wave</h3>

The area of the sound wave is calculated as follows;

A = 4\pi r^2\\\\A = 4 \pi \times (2)^2\\\\A = 50.272 \ m^2

<h3>Power generated</h3>

The sound power the person generated is calculated as follows;

P = I A\\\\P = 4.6\times 10^{-7} \ W/m^2 \ \ \times \ \ 50.272 \ m^2\\\\P = 2.313 \times 10^{-5} \ W

Learn more about intensity of sound here: brainly.com/question/4431819

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The wave shown below is produced in a rope.
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<h3>B) A dot vector drawn up and vector drawn down.</h3>

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What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne
irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance h= 1790 km = 1.790\times10^{6}\ m

Magnetic field B=4\times10^{-8}\ T

Mass of proton m=1.673\times10^{-21}\ Kg

Radius of earth R =6.38\times10^{6}\ m

Radius of orbit r=R+h

r=6.38\times10^{6}+1.790\times10^{6}

r=8170000\ m

We need to calculate the speed

Using formula of magnetic field

Bvq=\dfrac{mv^2}{r}

v=\dfrac{Bqr}{m}

Put the value into the formula

v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}

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Hence, The velocity is 31.25 m/s and direction is toward west.

6 0
3 years ago
An undersea research chamber is spherical with an external diameter of 5.50 m . The mass of the chamber, when occupied, is 87600
dmitriy555 [2]

Answer:

the buoyant force on the chamber is F = 7000460 N

Explanation:

the buoyant force on the chamber is equal to the weight of the displaced volume of sea water due to the presence of the chamber.

Since the chamber is completely covered by water, it displaces a volume equal to its spherical volume

mass of water displaced = density of seawater * volume displaced

m= d * V , V = 4/3π* Rext³

the buoyant force is the weight of this volume of seawater

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replacing values

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when occupied the tension force on the cable is

T = F buoyant - F weight of chamber  = 7000460 N - 87600 kg*9.8 m/s² = 6141980 N

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