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2 years ago

A 9-kg body has three times the kinetic energy of an 4-kg body. Calculate the ratio of the speeds of these bodies.

Physics

1 answer:

Neko [114]2 years ago

3 0

Answer:

speed 9/speed 4 = 2sqrt3 / 3

Explanation:

KE = 1/2 m v^2

for 9 kg = 3 * the 4 kg 1/2 (9)va^2 = <u> 3</u>* 1/2 (4) vb^2

va^2 / vb^2 = 3* 1/2 (4) / (1/2 *9)

6 / 4.5

va / vb = sqrt (6/4.5) = 2sqrt3 /3

ratio of the 9 speed to the speed of the 4 is 2 sqrt 3 /3

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It is Real,Virtual,The Same Size, Inverted

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3 years ago

A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the

insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

$\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}$

Req =3.03Ω , R1 =12.18Ω

$\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}$

$\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}$

$\frac{1}{R2}=0.2479$

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

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3 years ago

A car is traveling at a speed of 10m/s. A 0.5kg clump of mudis

CaHeK987 [17]

Answer:B

Explanation:

Given

speed of car $v=10 m/s$

mass of clump $m=0.5 kg$

Radius of car tire $r=0.2 m$

Since the tire is rotating about axle so a centripetal force is acting constantly on each particle towards the center of tire.

Centripetal force is given by

$F_c=\frac{mv^2}{r}$

where $m=mass\ of\ element$

$v=speed$

$r=distance\ from\ center$

$F_c=\frac{0.5\times 10^2}{0.2}$

$F_c=250\ N$ (inward)

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3 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 6.0 rev/s in 11.0 s. At th

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Answer

given,

ω₁ = 0 rev/s

ω₂ = 6 rev/s

t = 11 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

11 α = 6 - 0

= 0.545 rev/s²

The angular displacement

θ₁= ωi t + (1/2) α t²

θ₁= 0 + (1/2) (0.545)(11)^2

θ₁= 33 rev

case 2

ω₁ = 6 rev/s

ω₂ = 0 rev/s

t = 14 s

Using equation of rotational motion

The angular acceleration is

ωf - ωi = α t

14 α = 0 - 6

= - 0.428 rev/s²

The angular displacement

θ₂= ωi t + (1/2) α t²

θ₂= 6 x 14 + (1/2) (-0.428)(14)^2

θ₂= 42 rev

total revolution in 25 s is equal to

θ = θ₁ + θ₂

θ = 33 + 42

θ = 75 rev

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3 years ago

What is the energy stored between 2 Carbon nuclei that are 1.00 nm apart from each other? HINT: Carbon nuclei have 6 protons and

Andrei [34K]

Answer:

$A. 8.29\times 10^{-18}\ J$

Explanation:

Given that:

p = magnitude of charge on a proton = $1.6\times 10^{-19}\ C$

k = Boltzmann constant = $9\times 10^{9}\ Nm^2/C^2$

r = distance between the two carbon nuclei = 1.00 nm = $1.00\times 10^{-9}\ m$

Since a carbon nucleus contains 6 protons.

So, charge on a carbon nucleus is $q = 6p=6\times 1.6\times 10^{-19}\ C=9.6\times 10^{-19}\ C$

We know that the electric potential energy between two charges q and Q separated by a distance r is given by:

$U = \dfrac{kQq}{r}$

So, the potential energy between the two nuclei of carbon is as below:

$U= \dfrac{kqq}{r}\\\Rightarrow U = \dfrac{kq^2}{r}\\\Rightarrow U = \dfrac{9\times10^9\times (9.6\times 10^{-19})^2}{1.0\times 10^{-9}}\\\Rightarrow U =8.29\times 10^{-18}\ J$

Hence, the energy stored between two nuclei of carbon is $8.29\times 10^{-18}\ J$ .

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3 years ago