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djyliett [7]
4 years ago
14

A platinum sphere with radius 0.0199 m is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting o

n the sphere, and the sphere's apparent weight. The densities of platinum and mercury are 2.14×104 kg/m3 and 1.36×104 kg/m3, respectively.
Physics
1 answer:
zepelin [54]4 years ago
8 0

Answer:

Weight = 6.919 n

Buoyant force = 4.397 N

Apparent weight = 2.522 N

Explanation:

radius, r = 0.0199 m

density of platinum = 2.14 x 1064 kg/m^3

density of mercury = 1.36 x 1064 kg/m^3

Weight of sphere= volume of sphere x density of platinum x g

                            = 4/3 x 3.14 x (0.0199)^3 x 2.14 x 10^4 x 9.8 = 6.919 N

Buoyant force acting on sphere =  Volume of sphere x density of mercury x g

                            = 4/3 x 3.14 x (0.0199)^3 x 1.36 x 10^4 x 9.8 = 4.397 N

Apparent weight = True weight - Buoyant force = 6.919 - 4.397 = 2.522 N

         

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What is the total displacement of an ant that walks 2 meters west, 3 meters south, 4 meters east, and 1 meter
Ivan

(2m west) + (3m south) + (4m east) + (1m north) =

[ (2m west)+(4m east) ] + [ (3m south)+(1m north) ] =

[ (-2m east)+(4m east) ] + [ (3m south)-(1m south) ] =

(2m east) + (2m south)

Now, so far, we have the orthogonal (perpendicular) components of the displacement ... the North/South component and the East/West component.

To combine these, it's time for Pythagoras:

Displacement = √[ (2m)² + (2m)² ]

Displacement = √ (4m² + 4m²)

Diplacement = √8m²  

<em>Displacement =  2.83 meters Southeast</em>


8 0
4 years ago
A small remote-control car with a mass of 1.65 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow
Reil [10]

Answer:

N=63.69N

Explanation:

The normal force exerted on the car by the walls of the cylinder at the bottom of the vertical circle will be such that when substracted to the weight it must give the centripetal force, since at that point on the vertical F_{cp}=N-W=N-mg

We also know that the equation for the centripetal force is:

F_{cp}=ma_{cp}=\frac{mv^2}{r}

Mixing both equations we get:

N-mg=\frac{mv^2}{r}

N=mg+\frac{mv^2}{r}

Which for our values means:

N=(1.65Kg)(9.8m/s^2)+\frac{(1.65Kg)(12m/s)^2}{(5m)}=63.69N

8 0
3 years ago
Regarding the history of the universe, which of the following is true? Regarding the history of the universe, which of the follo
ale4655 [162]

Explanation:

1. The universe consisted of hydrogen and helium initially. This statement is true.

2. It is important to understand radioactive decay in order to understand this question, here's a good analogy:

A snake will shed it's skin, just as an atom will shoot off different parts of itself. It would be very difficult to force that snake to reenter it's skin once it sheds, just as it takes a lot of energy to force <em>fusion</em><em> </em>of atoms and the parts mentioned. In normal circumstances, nuclear decay is one-way.

3. The earth is a giant floating rock in space. It took many many years to gather a bunch of asteroids and dust to make this planet.

4. I shouldn't have to explain this one, it doesn't make much sense.

5 0
3 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and slip direction are at angles of 60° and 35°,
Vedmedyk [2.9K]

Answer:

19.5324 MPa

Explanation:

Information provided

Angle between the normal to the slip plane with tensile axis, \alpha=60^{o}&#10;

Angle by slip direction with tensile axis, \beta=35^{o}

Critical resolved shear stress, \tau_{c}=8 MPa

Applied stress \sigma=12 MPa

Shear stress at slip plane

\tau=\sigma cos\alpha cos\beta

\tau=12cos60^{o}cos35^{o}=4.915 MPa

\tau hence crystal won’t yield

Applied stress, \sigma for crystal to yield is given by

\sigma=\frac {\tau_{c}}{cos\alpha cos\beta}

\sigma=\frac {8}{cos60cos35}=19.53239342 MPa

\sigma=19.5324 MPa&#10;

7 0
3 years ago
A coil 4.20 cm radius, containing 500 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1
erma4kov [3.2K]

Explanation:

Given that,

Radius of the coil, r = 4.2 cm

Number of turns in the coil, N = 500

The magnetic field as a function of time is given by :

B=1.2\times 10^{-2}t+2.6\times 10^{-5}t^4

Resistance of the coil, R = 640 ohms

We need to find the magnitude of induced emf in the coil as a function of time. It is given by :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(NBA)}{dt}\\\\\epsilon=N\pi r^2\dfrac{-dB}{dt}\\\\\epsilon=N\pi r^2\times \dfrac{-d(1.2\times 10^{-2}t+2.6\times 10^{-5}t^4)}{dt}\\\\\epsilon=N\pi r^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=500\pi \times (4.2\times 10^{-2})^2\times (1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\\\\\epsilon=2.77(1.2\times 10^{-2}+10.4\times 10^{-5}t^3)\ V

Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
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