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4 years ago

15

Two objects, labeled A and B, are the same size. Object A has a density of 1.21 g/cm3. Object B has a density of 1.37 g/cm3. Bot

h are placed in a beaker of water. Which will float higher in the water?

2 answers:

Sedaia [141]4 years ago

4 0

Neither object will float in water.

They both have densities greater than 1.0 g/cm³,
so both will sink to the bottom of the beaker.

Like rocks.

deff fn [24]4 years ago

4 0

Answer:

None of the object float in water

Explanation:

When an object float in water than we can say that buoyancy force due to water will counter balance the weight of the object

As we can say that

$W_g = F_b$

so here if object is just floating or we can say that if object is in the state of sinking then we have

$\rho_{object} V g = \rho_{water} Vg$

so we have

$\rho_{object} = \rho_{water}$

so from above relation we can say that

i) $\rho_{object} > \rho_{water}$ then it will sink inside water

ii) $\rho_{object} < \rho_{water}$ then it will float in water

since we know density of water is 1 gm/cm^3 so here given both objects are having density more than this

so both the objects will sink inside water

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If an R = 1-kΩ resistor, a C = 1-μF capacitor, and an L = 0.2-H inductor are connected in series with a V = 150 sin (377t) volts

fgiga [73]

Answer

given,

R = 1-kΩ = 1000 Ω

C = 1-μF

L = 0.2-H

V = V_max sin( ω t)

comparing

V = 150 sin ( 377 t)

ω = 377

$\chi_c = \dfrac{1}{\omega C}$

$\chi_c = \dfrac{1}{377 \times 1 \times 10^{-6}}$

$\chi_c = 2652.5\Omega$

$\chi_L =377 \times 0.2$

$\chi_L =75.4\ \Omega$

Impedance,

$Z = \sqrt{R^2+(\chi_L-\chi_c)^2}$

$Z = \sqrt{1000^2+(75.4 -2652.5)^2}$

Z = 2764.3 Ω

now,

V_{max} = 150 V

$I_{max} = \dfrac{V}{Z}$

$I_{max} = \dfrac{150}{2764.3}$

$I_{max} = 0.0543$

$I_{max} = 54.3\ mA$

3 0

3 years ago

A wave traveling in the positive x-direction with a frequency of 50.0 Hz is shown in the figure below. Find the following values

Klio2033 [76]

Answer:

Explanation:

a. The amplitude is the measure of the height of the wave from the midline to the top of the wave or the midline to the bottom of the wave (called crests). The midline then divides the whole height in half. Thus, the amplitude of this wave is 9.0 cm.

b. Wavelength is measured from the highest point of one wave to the highest point of the next wave (or from the lowest point of one wave to the lowest point of the next wave, since they are the same). The wavelength of this wave then is 20.0 cm. or $\lambda=20.0cm$

c. The period, or T, of a wave is found in the equation

$f=\frac{1}{T}$ were f is the frequency of the wave. We were given the frequency, so we plug that in and solve for T:

$50.0=\frac{1}{T}$ so

$T=\frac{1}{50.0}$ and

T = .0200 seconds to the correct number of sig fig's (50.0 has 3 sig fig's in it)

d. The speed of the wave is found in the equation

$f=\frac{v}{\lambda}$ and since we already have the frequency and we solved for the wavelength already, filling in:

$50.0=\frac{v}{20.0}$ and

v = 50.0(20.0) so

v = 1.00 × 10³ m/s

And there you go!

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3 years ago

A proton with a speed of 3.000×105 m/s has a circular orbit just outside a uniformly charged sphere of radius 7.00 cm. What is t

Rama09 [41]

To solve this problem we will apply the principles of energy conservation. The kinetic energy in the object must be maintained and transformed into the potential electrostatic energy. Therefore mathematically

$KE = PE$

$\frac{1}{2} mv^2 = \frac{kq_1q_2}{r}$

Here,

m = mass (At this case of the proton)

v = Velocity

k = Coulomb's constant

$q_{1,2}$ = Charge of each object

r= Distance between them

Rearranging to find the second charge we have that

$q_2 = \frac{\frac{1}{2} mv^2 r}{kq_1}$

Replacing,

$q_2 = \frac{\frac{1}{2}(1.67*10^{-27})(3*10^5)^2(7*10^{-2})}{(9*10^9)(1.6*10^{-19})}$

$q_2 = 3.6531nC$

Therefore the charge on the sphere is 3.6531nC

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3 years ago

PLS HELP ME<br> I NEED HELP ASAP

Tanzania [10]

Answer:

In order to find average velocity we use the equation:
$avg = \frac{change \: in \: x}{change \: in \: time}$
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She did that in 4s
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3 years ago

At time t=0t=0 a proton is a distance of 0.360 mm from a very large insulating sheet of charge and is moving parallel to the she

insens350 [35]

Answer:

$1.34 * 10^{3}$ m/s

Explanation:

Parameters given:

distance of the proton form the insulating sheet = 0.360mm

speed of the proton, $v_{x}$ = 990m/s

Surface charge density, σ = 2.34 x $10^{-9}$ C/ $m^{2}$

We need to calculate the speed at time, t = 7.0 * $10^{-8}$ s.

We know that the proton is moving parallel to the sheet, hence, we can say it is moving in the x direction, with a speed $v_{x}$ on the axis.

The electric force acting on the proton moves in the y direction, so this means it is moving with velocity $v_{y}$ in the y axis.

Hence, the resultant velocity of the proton is given by:

$v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v_{x}$ = 990m/s from the question. We need to find $v_{y}$ and then the resultant velocity v.

Electric field is given in terms of surface charge density, σ as:

E = σ/ε0

where ε0 = permittivity of free space

=> $E = \frac{2.34 * 10^{-9} } {2 * 8.85418782 * 10^{-12} }$

E = 132 N/C

Electric Force, F is given in terms of Electric field:

F = eE

where e = electronic charge

=> F = ma = eE

∴ a = eE/m

where

a = acceleration of the proton

m = mass of proton

$a = \frac{1.60 * 10^{-19} * 132}{1.672 * 10^{-27} }$

a = 1.3 * $10^{10}$ m/ $s^{2}$

Therefore, at time, t = 7.0 * $10^{-8}$ , we can use one of the equations of linear motion to find the velocity in the y axis:

$a = \frac{v_{y} - v_{0}}{t} \\\\=> v_{y} = v_{0} + at$

$v_{y}$ = 0 + (1.3 * $10^{10}$ * 7.0 * $10^{-8}$ )

$v_{y}$ = 910 m/s

∴ $v = \sqrt{v_{x} ^{2} + v_{y} ^{2}}$

$v = \sqrt{990^{2} + 910^{2} }$

$v = \sqrt{1808200}$

v = 1344.69 m/s = $1.34 * 10^{3}$ m/s

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4 years ago