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SCORPION-xisa [38]
3 years ago
9

4. A steel cable spanning a river is 220.000 m long when the temperature is 30.°C.

Physics
1 answer:
romanna [79]3 years ago
5 0

Answer:

219.9208  m

Explanation:

The new length is given by

New length= Original length *(1-Temperature change*coefficient of

thermal expansion of steel)

Here, the change in temperature is 30^{\circ}-0^{\circ}= 30^{\circ}

New length= 220.000(1-30*12x10^{-6}=219.9208m

Therefore, the new length will be 219.9208 m

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this event is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the oc
MissTica

Answer: Tsunami.

Explanation :

Tsunami is the result of a sudden vertical offset in the ocean floor and is most often the result of plate movement on the ocean floor.

Tsunami is caused due to the displacement of a large volume of water like in an ocean. It consists of a series of waves. It destroys coastlines and coastal settlements. It is also known as a tidal wave.

So, the correct option is (b) Tsunami.

6 0
3 years ago
Can someone help me with this please?​
Nikolay [14]

Answer:

Explanation:

umm... try 30

4 0
3 years ago
A diffraction grating is illuminated simultaneously with red light of wavelength 670 nm and light of an unknown wavelength. The
marusya05 [52]

Answer:

dsin∅ = m× λ

so, dsin∅red = 3(670nm)

also, dsin∅? =5λ?

however ,if they overlap then dsin∅red = dsin∅?

3(670nm) /5 =402nm

∴λ = 402nm

Explanation:

4 0
3 years ago
A nonconducting sphere has radius R = 1.29 cm and uniformly distributed charge q = +3.83 fC. Take the electric potential at the
zalisa [80]

Answer:

a) -2.516 × 10⁻⁴ V

b) -1.33 × 10⁻³ V

Explanation:

The electric field inside the sphere can be expressed as:

E= \frac{kqr}{R^3}

The potential at a distance can be represented as:

V(r) - V(0) = -\int\limits^r_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) = [\frac{qr^2}{8 \pi E_0R^3 }]₀

V(r) =   -[\frac{qr^2}{8 \pi E_0R^3 }]₀

Given that:

q = +3.83 fc = 3.83 × 10⁻¹⁵ C

r = 0.56 cm

 = 0.56 × 10⁻² m

R = 1.29 cm

  =  1.29 × 10⁻² m

E₀ = 8.85 × 10⁻¹² F/m

Substituting our values; we have:

V(r) = -\frac{(3.83*10^{-15}C)(0.560*10^{-2}m)^2}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)^3}

V(r) = -2.15  × 10⁻⁴ V

The difference between the radial distance  and center can be expressed as:

V(r) - V(0) = -\int\limits^R_0 {\frac{kqr}{R^3} } \, dr^2

V(r) - V(0) =  [\frac{qr^2}{8 \pi E_0R^3 }]^R

V(r) = -\frac{qR^2}{8 \pi E_0R^3 }

V(r) = -\frac{q}{8 \pi E_0R }

V(r) = -\frac{(3.83*10^{-15}C)}{8 \pi (8.85*10^{-12}F/m)(1.29*10^{-2}m)}

V(r) = -0.00133

V(r) = - 1.33 × 10⁻³ V

8 0
3 years ago
What has research determined about the orbit of an electron around a nucleus?
sergeinik [125]
The one that research has determined about the orbit of an electron around nucleus is : Each sub-level electron type has a unique path where it will likely to be found
Here are the sub levels of an electron :
-sub level s, maximum number of 2 electrons
- sub level p, maximum number of 6 electrons
- sub level d, maximum number of 10 electrons
- sub level f, maximum number of 14 electrons
8 0
3 years ago
Read 2 more answers
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