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Thepotemich [5.8K]
3 years ago
14

Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is five times greater. Find its r

adius. A. 22.5 cm. B. 6.36 cm. C. 0.90 cm. D. 7.69 c
Physics
2 answers:
Sonja [21]3 years ago
4 0
Vol of sphere = 4/3 pi r^2.density of sphere = mass/volume.mass = densityxvolumesphere 1. mass = density x 4/3 pi 4.5^2sphere 2 5mass = density x 4/3 pi r^25=4/3 pi r^2 divided by 4/3 pi 4.5^25=r^2 divided by  4.5^25x4.5^2=r^2root(5x4.5^2)=r4.5 root 5 = r
zloy xaker [14]3 years ago
4 0

Answer:

D) 7.69 cm

Explanation:

As we know that both the sphere must have same density

So the volume of first sphere

V = \frac{4}{3}\pi r^3

V = \frac{4}{3}\pi(4.50)^3

now the mass of this sphere is given as

m = \rho V

m = \rho(\frac{4}{3}\pi (4.50)^3)

now we know that mass of other sphere is 5 times more than this

so we will have

m_1 = 5 m

\rho(\frac{4}{3}\pi r^3) = 5\rho(\frac{4}{3}\pi (4.50)^3)

r^3 = 5(4.50)^3

r = 7.69 cm

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2 years ago
what is required to bring about a phase change 1.an increase or decrease in pressure 2.an increase or decrease in energy 3.an in
frozen [14]

Answer

Correct answers are 1.an increase or decrease in pressure 2.an increase or decrease in energy

Explanation

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8 0
3 years ago
Read 2 more answers
The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

Also;

\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

where;

\delta = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

k = 104.82\ \  N/m

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

\frac{1.53(0.6+0.6)^2}{3} \omega_3^2  + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0

0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

\omega _3 =1.70 \ rad/s

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

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Venus would be an inner planet!
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3 years ago
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