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4 years ago

14

A ________ stream pattern develops on lands underlain by tilted or folded, alternating hard and soft, sedimentary strata.

1 answer:

Mariana [72]4 years ago

4 0

The answer in this question is trellis in which define in this statement as the stream pattern which develops on lands underlain by the tilted or folded, alternating hard and soft and sedimentary strata is trellis. Trellis in architecture is an archintectural structure that is usually made from an open framework or lattice of interwoven or intersecting pieces of wood, bamboo or metal.

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Which of the following statements describe the transfer of energy a. Collision of atoms causing nuclear reactions B. A car speed

oksano4ka [1.4K]

I think the answer would be A.

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3 years ago

Swimmers at the beach are tanning on towels. Which method of heat transfer is responsible for their tan? *

Free_Kalibri [48]

Answer:

ratiation

Explanation:

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3 years ago

Read 2 more answers

A property unique to a conducting substance that determines its resistance is called:

elena-s [515]

Answer:

B. Resistivity

Explanation:

Resistance offered by a substance of unit area per unit length.

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3 years ago

A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of

Vesna [10]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

$\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}$

Where, $\beta=\dfrac{c}{v}$

$\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}$

$\lambda_{o}=115\ nm$

Hence, The observer detects light of wavelength is 115 nm.

8 0

4 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago