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lana [24]
3 years ago
9

Please help me to solve this and give me the summary of answer

Physics
1 answer:
Neko [114]3 years ago
7 0
Using the principle of floatation.

u = w............(a)

Upthrust of fluid is equal to the weight of the object.

Let the volume of the wood be V.

The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V

Formula for upthrust, u = vdg

where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity

weight, w = mg
where m = mass
g = acceleration due to gravity

From (a)

                     u = w

                 vdg =  mg      Cancel out g

                   vd  =  m
 
The v  is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.

d is density of fluid which is water in this case, 1000 kg/m³

         0.2V * 1000 =  m

           200V =  m

Hence the mass of the object is  200V  kg.

But Density of solid =  Mass of solid / Volume of solid

                                 =    200V / V

                                 =    200 kg/m³

Density of solid = 200 kg/m³     
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Find the electric energy density between the plates of a 225-μF parallel-plate capacitor. The potential difference between the p
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Answer:

Energy density will be 14.73 J/m^3

Explanation:

We have given capacitance C=225\mu F=225\times 10^{-6}F

Potential difference between the plates = 365 V

Plate separation d = 0.200 mm 0.2\times 10^{-3}m

We know that there is relation between electric field and potential

E=\frac{V}{d}, here E is electric field, V is potential and d is separation between the plates

So E=\frac{V}{d}=\frac{365}{0.2\times 10^{-3}}=1825000N/C

Energy density is given by E=\frac{1}{2}\varepsilon _0E^2=\frac{1}{2}\times 8.85\times 10^{-12}\times (1.825\times 10^6)^2=14.73J/m^3

5 0
3 years ago
What is the definition of physics
Stella [2.4K]

Answer:

the branch of science concerned with the nature and properties of matter and energy. The subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms.

3 0
2 years ago
I just need number 2
il63 [147K]

We will apply the conservation of linear momentum to answer this question.

Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Note that v₁ and v₁' is the velocity of m₁ before and after the collision.

Let's choose m₁ and v₁ to represent the bullet's mass and velocity.

m₂ and v₂ represents the wood block's mass and velocity.

The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'

m₁v₁ + m₂v₂ = m₁v' + m₂v'

m₁v₁ + m₂v₂ = (m₁+m₂)v'

Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.

m₁v₁ = (m₁+m₂)v'

Here are the given values:

m₁ = 0.005kg

v₁ = 500m/s

m₂ = 5kg

Plug in the values and solve for v'

0.005×500 = (0.005+5)v'

v' = 0.4995m/s

v' ≅ 0.5m/s

4 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
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solution:

y = v0t + ½at²

1150 = 79t + ½3.9t²

0 = 3.9t² + 158t - 2300

from quadratic equations and eliminating the negative answer

t = (-158 + v158² -4(3.9)(-2300)) / 2(3.9)

t = 11.37 s to engine cut-off

the velocity at that time is

v = v0 + at

v = 79 + 3.9(11.37)

v = 123.3 m/s

it rises for an additional time

v = gt

t = v/g

t = 123.3 / 9.8

t = 12.59 s

gaining more altitude

y = ½vt

y = 123.3(12.59) /2

y = 776 m

for a peak height of

y = 776 + 1150


5 0
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Answer:

c

Explanation:

the loud noise can reduce the quality of the analog signal

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