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olga55 [171]
3 years ago
6

How know that rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the arkose sandstone

Physics
2 answers:
svp [43]3 years ago
7 0
The rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the Arkose sandstone because the Amadeus basin were made up of marine and non-marine sedimentary rocks which are softer compared to quarts which make up mostly the Arkose sandstone.
Serhud [2]3 years ago
6 0
Hello there.

<span>How know that rocks formed from the seafloor sediments deposited in the Amadeus basin were softer than the arkose sandstone
</span>
Amadeus basin were made up of marine and non-marine sedimentary rocks which are softer compared to quarts which make up mostly the Arkose sandstone.
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A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe
Eduardwww [97]

Answer:

Final temperature, T_f=21.85^{\circ}

Explanation:

Given that,

Mass of silver ring, m = 4 g

Initial temperature, T_i=41^{\circ}C

Heat released, Q = -18 J (as heat is released)

Specific heat capacity of silver, c=0.235\ J/g\ C

To find,

Final temperature

Solution,

The expression for the specific heat is given by :

Q=mc\Delta T

Q=mc(T_f-T_i)

T_f=\dfrac{-Q}{mc}+T_i

T_f=\dfrac{-18}{4\times 0.235}+41

T_f=21.85^{\circ}

So, the final temperature of silver is 21.85 degrees Celsius.

5 0
3 years ago
In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The fron
slega [8]

Answer:

a)   v_average = 11 m / s, b)  t = 0.0627 s

, c)    F = 7.37 10⁵ N

, d)   F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

The average speed is

           v_average = (v + v₀) / 2

           v_average = (22 + 0) / 2

           v_average = 11 m / s

b) The average time

          v = v₀ - a t

          t = v₀ / a

          t = 22 / 350.72

          t = 0.0627 s

c) The force can be found with Newton's second law

             F = m a

             F = 2100 350.72

             F = 7.37 10⁵ N

.d) the ratio of this force to weight

             F / W = 7.37 10⁵ / (2100 9.8)

             F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

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2 years ago
What is universal laws of gravitation ​
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Answer:

Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

Explanation:

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3 years ago
What happens to gravitational potential energy as a roller coaster moves down a hill?
Kaylis [27]
It is converted to kinetic energy.
5 0
3 years ago
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