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Ivahew [28]
3 years ago
9

A platinum sphere with radius 1.21 cm is totally immersed in mercury. Find the weight of the sphere, the buoyant force acting on

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:
TEA [102]3 years ago
5 0

Answer:

1.556 N , 0.989 N, 0.567 N

Explanation:

Radius of sphere, r = 1.21 cm = 0.0121 m

density of platinum , d = 2.14 x 10^4 kg/m^3

density of mercury, d' = 13.6 x 10^3 kg/m^3

Volume of sphere, \frac{4}{3}\pi \times r^{3}= \frac{4}{3}\times 1.34\times \left (0.0121  \right )^{3}

V = 7.42 x 10^-6 m^3

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

W = V x d x g = 7.42 x 10^-6 x 2.14 x 10^4 x 9.8 = 1.556 N

Buoyant force, B = Volume x density of mercury x gravity

B = 7.42 x 10^-6 x 13.6 x 10^3 x 9.8 = 0.989 N

Apparent weight = True weight - Buoyant force

Apparent weight = 1.556 - 0.989 = 0.567 N

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Most of the fusion occurs in the core.

In stars more massive than the sun, fusion continues through Deuterium, Carbon, and finally reaching iron/nickel.

Up to this point, the fusion reaction was endothermic, which means that the energy expended to produce the fusion reaction was exceeded by the energy produced in the reaction.

Fusion past iron is exothermic, and therefore the star will be able to survive by fusing elements heavier than iron.

After the core is almost entirely iron, the star is no longer in the Main Sequence.

So, fusion in stars more massive than the sun continue fusing until the core is almost entirely <em>iron</em>.
3 0
4 years ago
Could someone explain to me how to got the answer B, thank you very much​
Mnenie [13.5K]

Answer:

since -6 lasted for 5 seconds, multiplying both would result in -30

3 lasted for 10 seconds, so multiplying both would give +30

average = ( 30 + (-30) ) / 2

30 -30 is already equal to zero, so the answer should be 0

4 0
3 years ago
As you sit in a chair, in what direction (or directions) do you exert an action force?
enyata [817]
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5 0
4 years ago
Read 2 more answers
A car moves round a circular track of radius 0.3m of two revolution per/sec find its angular velocity.
Pie

Answer:

the angular velocity of the car is 12.568 rad/s.

Explanation:

Given;

radius of the circular track, r = 0.3 m

number of revolutions  per second made by the car, ω = 2 rev/s

The angular velocity of the car in radian per second is calculated as;

From the given data, we convert the angular velocity in revolution per second to radian per second.

\omega = 2 \ \frac{rev}{s} \times \frac{2\pi \ rad}{1 \ rev} = 4\pi \ rad/s = 12.568 \ rad/s

Therefore, the angular velocity of the car is 12.568 rad/s.

4 0
3 years ago
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
3 years ago
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