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3 years ago

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A metal cylinder measures 3.0 cm in diameter, is 12.0 cm high, and has a mass of 750 g. The cylinder hangs from a string attache

d to a scale that measures weight in Newtons (N). The cylinder is lowered into a cylindrical beaker of water until it is submerged completely. The diameter of the beaker is 10.0 cm. (a) By how much does the height of the water in the beaker rise when the cylinder is submerged? (b) What weight does the scale read after the cylinder is submerged?

1 answer:

kvasek [131]3 years ago

3 0

Answer with Explanation:

We are given that

Diameter of cylinder, $d_1=$ 3 cm

Radius, $r_1=\frac{d_1}{2}=\frac{3}{2}=1.5cm=1.5\times 10^{-2} m$

1m=100 cm

Height,h=12 cm= $\frac{12}{100}=0.12 m$

Mass,m=750 g= $\frac{750}{1000}=0.75kg$

1kg=1000 g

Diameter of beaker, $d_2=10 cm$

Radius of beaker, $r_2=\frac{d_2}{2}=\frac{10}{2}=5cm$

A. $\pi r^2_2h_2=\pi r^2_1 h_1$

$h_2=\frac{r^2_1}{r^2_2}h$

$h_2=\frac{(1.5)^2}{(5)^2}\times 12=1.08 cm$

Hence, the height of the water in the beaker rises when the cylinder is submerged=1.08 cm

B.Weight read on the scale after the cylinder is submerged

$w=mg-\rho A_1h_1g$

Where $g=9.8 m/s^2$

$A_1=\pi r^2_1$

Density of water= $\rho=1000kg/m^3$

$w=0.75\times 9.8-1000\times \pi(1.5\times 10^{-2})^2\times 0.12\times 9.8$

$w=6.52 N$

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Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

kramer

Answer:

E = 440816.32 N/C

Explanation:

Given data:

Three point charge of charge equal to +3.0 micro coulomb

fourth point charge = - 3.0 micro coulomb

side of square = 0.50 m

$K =1/4 \pi \epsilon_0 = 8.99 \times 10^9$ N.m^2/c^2

Due to having equal charge on center of square, 2 charge produce equal electric field at center and other two also produce electric field at center of same value

So we have

$E_1 + E_3 = 0$

$E =E_2 + E_4$

$E = 2 E_2$

[ $E_2 =\frac{2\times k \times q}{r^2}$

[ $r= \frac{(0.5^2 + 0.5^2)^2}{2} = 0.35 m$ ]

plugging all value

$E = 2 E_2$

$E = 2 E_2 =\frac{2\times k \times q}{r^2}$

$E = \frac{2 \times 8.99 \times 10^93\times 10^{-6}}{0.35^2}$

E = 440816.32 N/C

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Answer:

evolutionary theory

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The evidence is overwhelming that the Grand Canyon was dug over a span of millions of years by the erosive power of the Colorado River and that river’s tributary streams.

This evidence does not support catastrophic theory but it does support evolutionary theory, because it was gradual process and took place overtime. It did not happen in a catastrophic manner.

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2 years ago

The electric field 2.5 mm from a uniform sheet of charge is σ=800, NC. How much charge is contained in a 5.0x5.0 cm section of t

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Answer:

The charge is $2.75\times10^{-13}\ C$

Explanation:

Given that,

Distance = 2.5 mm

Electric field = 800 NC

Length $L=5.0\times5.0\times10^{-4}\ m$

We need to calculate the linear charge density

Using formula of linear charge density

$E=\dfrac{2k\lambda}{r}$

$\lambda=\dfrac{Er}{2k}$

Put the value into the formula

$\lambda=\dfrac{800\times2.5\times10^{-3}}{2\times9\times10^{9}}$

$\lambda=1.1\times10^{-10}\ C/m$

We need to calculate the charge

Using formula of charge

$Q=\lambda\timesL$

Put the value into the formula

$Q=1.1\times10^{-10}\times(5.0\times5.0\times10^{-4})$

$Q=2.75\times10^{-13}\ C$

Hence, The charge is $2.75\times10^{-13}\ C$

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2 years ago

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if i am changing velocity, i must also have <u>acceleration</u> and a net <u>force</u>

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2 years ago

A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

DochEvi [55]

Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

d) $K.E=0.096J$

e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

a)

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

Therefore

$V_m=A\omega\\\\V_m=8.1*0.1$

$V_m=0.81m/s$

c)

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

$a_m=\omega^2A$

$a_m=8.1^2*0.1$

$a_m=6.561m/s^2$

d)

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

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

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$K.E=0.096J$

e)

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

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Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

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f)

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

Where

'= signify order of differentiation

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3 years ago