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ch4aika [34]
3 years ago
5

A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br

eak when the tension reached how many newtons
Physics
1 answer:
serg [7]3 years ago
7 0

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

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Kepler-62e is an exoplanet that orbits within the habitable zone around its parent star. The planet has a mass that is 3.57 time
skad [1K]

Answer:

g' = 13.5 m/s²

Explanation:

The acceleration due to gravity on surface of earth is given by the formula:

g = GMe/Re²   --------------- euation 1

where,

g = acceleration due to gravity on surface of earth

G = Universal Gravitational Constant

Me = Mass of Earth

Re = Radius of Earth

Now, the the acceleration due to gravity on the surface of Kepler-62e is:

g' = GM'/R'²   --------------- euation 1

where,

g' = acceleration due to gravity on surface of Kepler-62e

G = Universal Gravitational Constant

M' = Mass of Kepler-62e = 3.57 Me

R' = Radius of Kepler-62e = 1.61 Re

Therefore,

g' = G(3.57 Me)/(1.61 Re)²

g' = 1.38 GMe/Re²

using equation 1:

g' = 1.38 g

where,

g = 9.8 m/s²

Therefore,

g' = 1.38(9.8 m/s²)

<u>g' = 13.5 m/s²</u>

6 0
3 years ago
Suppose that when you move the north pole of a bar magnetic toward a coil it induces a positive current in the coil. The strengt
dolphi86 [110]

Answer:

<em>a) A positive current will be induced in the coil</em>

Explanation:

Electromagnetic induction is the induction of an electric field on a conductor due to a changing magnetic field flux. The change in the flux can be by moving the magnet relative to the conductor, or by changing the intensity of the magnetic field of the magnet. In the case of this electromagnets<em>, the gradual increase in the the electromagnet's field strength will cause a flux change, which will in turn induce an electric current on the coil.</em>

According to Lenz law, the induced current acts in such a way as to negate the motion or action that is producing it. <em>A positive current will be induced on the coil so as to repel any form of attraction between the north pole of the electromagnet and the coil</em>. This law obeys the law of conservation of energy, since work has to be done to move the move them closer to themselves.

3 0
3 years ago
The speed of light is about 300,000 km/s (3×105km/s). If Jupiter is 0.72 light hours from the Sun, how far is this?
Daniel [21]

Answer:

7.77 x 10⁸ Km

Explanation:

given,

Speed of light = 3 x 10⁵ Km/s

                        = 3 x 10⁸ m/s

1 light hour = 1.079 x 10⁹ Km

now,

0.72 light hour = 0.72 x 1.079 x 10⁹ Km

                         = 0.7769 x 10⁹ Km

                         = 7.77 x 10⁸ Km

The Jupiter is 7.77 x 10⁸ Km far from sun.

7 0
4 years ago
Read 2 more answers
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

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4 years ago
Name and describe the six layers of the sun, in order from innermost to outermost.
AlexFokin [52]
1.) core- the central region where nuclear reactions consume hydrogen to form helium

2.) radiative zone- the section of solar interior between the innermost core and the outer convective zone.

3.) convection zone- where light (photons) is created.

4.) photosphere- the visible surface of the sun that we are most familiar with

5.) chromosphere- second most outer layer of the sun. due to low density it is relatively transparent and only apparent during eclipses.

6.) corona- an aura of plasma that surrounds the sun and other stars but is usually hidden by the bright light of the sun’s surface.
4 0
3 years ago
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