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2 years ago

Who Designed the one-piece wrought iron plow.

Physics

1 answer:

marissa [1.9K]2 years ago

8 0

Answer:

Jethro Wood is the one who designed the one-piece wrought iron plow.

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Suatu gas memiliki volume 2 m³ dipanaskan dengan kondisi isobarik sehingga volumenya menjadi 5,5 m³ jika tekanan 4 atmosfer hitu

Ivan

Answer:

Sorry if wrong.............

5 0

2 years ago

A circular 10-turn coil with a radius of 5.0 cm carries a current of 5 A. It lies in the xy plane in a uniform magnetic field =

Tju [1.3M]

Answer:

U = – 0.12J

Explanation:

Given N = 10 turns, I = 5A, r = 5×10-²m

B^ = 0.05 T iˆ+ 0.3 T kˆ

Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T

Area = πr² = π(5×10-²)² = 7.85×10-³m²

Magnetic moment μ = NIA

μ = 10×5×7.85×10-³ = 0.3925Am²

U = -μ•B = –0.3925×0.304 = –0.12J

The sign is negative because the magnetic moment is aligned with the magnetic field.

3 0

2 years ago

What is a literature review?<br>

almond37 [142]

Answer: A literature review consists of an overview, a summary, and an evaluation (“critique”) of the current state of knowledge about a specific area of research.

Explanation:

7 0

2 years ago

A block of wood of mass 24 kg floats on water. The volume of the block below the surface of the water and the density of the woo

Slav-nsk [51]

Answer:

0.024m^3

Explanation:

=======

Answer:

=======

Given:-

Mass of the block of wood = 24 kg

Volume of wood = 0.032 m^3

Density of water = 1000kg/m^3

Now,

Density of wood is given by,

\frac{m}{v} = \frac{24}{0.032} \\

\frac{m}{v} = 750 \: kg/m ^{3}

Therefore,

The density of wood is 750kg/m^3

By principle of floatation,

Mass \:of\: wood = Mass\: of\: liquid \:displaced

Therefore,

Mass of liquid displaced = 24kg

Volume of liquid displaced (v),

\frac{m}{v} = \frac{24}{1000} \\

\frac{m}{v} = 0.24m ^{3}

Now,

Since the volume of the wood is equal to the volume of water displaced, it is 0.024m^3

=====

Note:

=====

=> The volume of the wood below the water surface is the volume of water displaced.

=> Buoyant\: force = Weight\: of\: the \:displaced\: water.

8 0

3 years ago

(1 point) A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 new

Nonamiya [84]

Answer:

x(t)=0.337sin((5.929t)

Explanation:

A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.

Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation

$m \frac{d^{2}x}{dt^{2}} +kx=0$

Definition of parameters

m=mass 3kg

k=force constant

e=extension ,m

ω =angular frequency

k=90/1.6=56.25N/m

ω^2=k/m= 56.25/1.6

ω^2=35.15625

ω=5.929

General solution will be

$x(t)=c1cos(ωt)+c2Sin(ωt)$

$x(t)=c1cos(5.929t)+c2Sin(5.929t)$

differentiating x(t)

dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)

when x(0)=0, gives c1=0

dx(t0)=2m/s gives c2=0.337

Therefore, the position of the mass after t seconds is

x(t)=0.337sin((5.929t)

6 0

2 years ago