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IceJOKER [234]
1 year ago
7

A beaker of mass 1.2 kg containing 2.5 kg of water rests on a scale. A 3.8 kg block of a metallic alloy of density 3300 kg/m3 is

suspended from a spring scale and is submerged in the water of density 1000 kg/m? as shown in the figure. a) What does the hanging scale read? acceleration of gravity is 9.8 m/s?Answer in units of N.b) what does the lower scale read? Answer in units of N.

Physics
1 answer:
Sphinxa [80]1 year ago
4 0

ANSWER:

a) 25.97 N

b) 47.53 N

STEP-BY-STEP EXPLANATION:

Given:

Beaker mass = 1.2 kg

Water mass = 2.5 kg

Water density = 1000 kg/m^3

Block mass = 3.8 kg

Block density = 3300 kg/m^3

a)

The first thing is to calculate the volume of the block, like this:

\begin{gathered} d=\frac{m}{V} \\ V=\frac{m}{d} \\ \text{ we replacing} \\ V=\frac{3.8}{3300} \\ V=0.00115m^3 \end{gathered}

Mass of water displaced by the block is:

\begin{gathered} d=\frac{m}{V} \\ m=d\cdot V \\ m=1000\cdot0.00115 \\ m=1.15kg \end{gathered}

The block will receive a push from the water equal to the weight of the water displaced by the block, or the effective weight of the block will be reduced by the same amount:

\begin{gathered} W=(m_b-m)\cdot g \\ \text{ we replacing} \\ W=(3.8-1.15)\cdot9.8 \\ W=25.97\text{ N} \end{gathered}

Therefore, 25.97 N is the reading on the hanging scale.

b)

The bottom scale will gain by the same amount (1.15 kg). Therefore, the totalweight on the bottom scale is:

\begin{gathered} W=(1.2+2.5+1.15)\cdot9.8 \\ W=4.85\cdot9.8 \\ W=47.53\text{ N} \end{gathered}

Therefore, 47.53 N is the reading on the lower scale.

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Explanation:

Given that,

Radius of the disk, r = 0.25 m

Mass, m = 45.2 kg

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Angle made by the ramp with horizontal, \theta=17^{\circ}

Solution,

As the disk starts from rest from the top of the ramp, the potential energy is equal to the sum of translational kinetic energy and the rotational kinetic energy or by using the law of conservation of energy as :

(a) mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2

h is the height of the ramp

sin\theta=\dfrac{h}{5.4}

h=sin(17)\times 5.4=1.57\ m

v is the speed of the disk's center

I is the moment of inertia of the disk,

I=\dfrac{1}{2}mr^2

\omega=\dfrac{v}{r}

mgh=\dfrac{1}{2}mv^2+\dfrac{1}{2}\dfrac{1}{2}mr^2\times (\dfrac{v}{r})^2

gh=\dfrac{1}{2}v^2+\dfrac{1}{4}v^2

gh=\dfrac{3}{4}v^2

9.8\times 1.57=\dfrac{3}{4}v^2

v = 4.52 m/s

(b) At the bottom of the ramp, the angular speed of the disk is given by :

\omega=\dfrac{v}{r}

\omega=\dfrac{4.52}{0.25}

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At its closest approach, a moon comes within 200,000 km of the planet it orbits. At that point, the moon is 300,000 km from the
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Answer:

s₁ = 240,000 km

Explanation:

The distance between both the focuses f₁ and f₂ will be the sum of distances of the moon from each focus at a given point. Therefore,

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Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 51.5 A. The resistanc
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Answer:

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a).

ζcu=1.7x10^{-8}Ωm

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d).

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Answer:

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Explanation:

Hope this helps!! :)

4 0
3 years ago
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