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Rzqust [24]
2 years ago
13

A ball with a volume of 3 cm3 and a density of 9.0 g/cm3 increases its velocity from 2 m/s to 6 m/s over a 12

Physics
1 answer:
V125BC [204]2 years ago
4 0

Answer:

M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm

a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2     the acceleration

F = M a = 27 gm * 1/3 m/s^2 = 9 dynes      net force applied

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Two bumper cars collide into each other and each car jolts backwards. which one of Newton's laws is this?
LekaFEV [45]
I think it would be Newton’s second law
7 0
2 years ago
A long, straight, cylindrical wire of radius R carries a current uniformly distributed over its cross section.
Mice21 [21]

Answer:

Explanation:

We shall solve this question with the help of Ampere's circuital law.

Ampere's ,law

∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire

we shall find magnetic field at distance x . current enclosed in the area of circle of radius x

=  I x π x²  / π R²

= I x²  /  R²

B x 2π x = μ₀  x current enclosed

B x 2π x = μ₀  x  I x²  /  R²

B =  μ₀   I x  / 2π R²

Maximum magnetic B₀ field  will be when x = R

B₀ = μ₀I   / 2π R

Given

B = B₀ / 3

μ₀   I x  / 2π R² = μ₀I   / 2π R x 3

x = R / 3

b ) The largest value of magnetic field is on the surface of wire

B₀ = μ₀I   / 2π R

At distance x outside , let magnetic field be B

Applying Ampere's circuital law

∫ B dl = μ₀ I

B x 2π x = μ₀ I

B = μ₀ I / 2π x

Given B = B₀ / 3

μ₀ I / 2π x = μ₀I   / 2π R x 3

x = 3R .

3 0
3 years ago
Which of the following is a good conductor of heat? 1. water 2. glass 3. wood 4. mercury​
trapecia [35]

Answer:

So the answer would be water based on the evidence shown below.

Explanation:

Mercury is a poor conductor of heat but good for electricity, water is a good conductor of heat but a poor conductor of electricity, wood is a poor conductor of heat and electricity, and glass is probably the worst conductor of heat.

7 0
3 years ago
What total mass must be converted into energy
Eduardwww [97]

This question apparently wants you to get comfortable
with  E = m c² .  But I must say, this question is a lame
way to do it.

c = 3 x 10⁸ m/s
                                                    E = m c²

                           1.03 x 10⁻¹³ joule  =  (m) (3 x 10⁸ m/s)²

Divide each side by (3 x 10⁸ m/s)²:

                         Mass = (1.03 x 10⁻¹³ joule) / (9 x 10¹⁶ m²/s²)

                                   =  (1.03 / 9) x (10⁻¹³ ⁻ ¹⁶) (kg)

                                   =        1.144 x 10⁻³⁰  kg .    (choice-1)

This is roughly the mass of (1 and 1/4) electrons, so it seems
that it could never happen in nature.  The question is just an
exercise in arithmetic, and not a particularly interesting one.
______________________________________

Something like this could have been much more impressive:
 
The Braidwood Nuclear Power Generating Station in northeastern
Ilinois USA serves Chicago and northern Illinois with electricity.
<span>The station has two pressurized water reactors, which can generate
a net total of 2,242 megawatts at full capacity, making it the largest
nuclear plant in the state.
If the Braidwood plant were able to completely convert mass
to energy, how much mass would it need to convert in order
to provide the total electrical energy that it generates in a year,
operating at full capacity ?

Energy = (2,242 x 10⁶ joule/sec) x (86,400 sec/day) x (365 da/yr)

             =  (2,242 x 10⁶ x 86,400 x 365) joules

             =          7.0704 x 10¹⁶ joules .

How much converted mass is that ?

                                           E  =  m c²

Divide each side by  c² :    Mass  =  E / c² .
c = 3 x 10⁸ m/s

              Mass = (7.0704 x 10¹⁶ joules) / (9 x 10¹⁶ m²/s²)

                        =        0.786 kilogram ! ! !

THAT should impress us !  If I've done the arithmetic correctly,
then roughly  (1 pound  11.7 ounces) of mass, if completely
converted to energy, would provide all the energy generated
by the largest nuclear power plant in Illinois, operating at max
capacity for a year !

</span>
7 0
3 years ago
Read 2 more answers
A cubical block of iron 10 cm on each side is floating on mercury in a vessel. (i) What is the height of the block above the mer
Gre4nikov [31]

Answer:

i 5.3 cm ii. 72 cm

Explanation:

i

We know upthrust on iron = weight of mercury displaced

To balance, the weight of iron = weight of mercury displaced . So

ρ₁V₁g = ρ₂V₂g

ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?

V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³

So, the height of iron above the mercury is h = V₂/area of base iron block

= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm

ρ₁V₁g = ρ₂V₂g

ii

ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?

V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ =  7200 cm³

So, the height of column of water is h = V₃/area of base iron block

= 7200 cm³/10² cm² = 72 cm

7 0
3 years ago
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