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Natali [406]
3 years ago
7

If you cut a piece of wood in half, each half would have *

Physics
1 answer:
Anon25 [30]3 years ago
7 0

Answer:

c

Explanation:

let's assume that you are splitting the piece of wood horizontally. This would not change the density of the wood at all, only the length. I rest my case. Ur awesome and have an amazing day.

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A dancer lifts his partner above his head with an acceleration of 2.5 m/s squared the dancer. Exerts a force of 200N what is the
elena-s [515]
F = ma
F = 200N, a = 2.5m/s^2

200N = (2.5)m
m = 80kg
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Roseanne heated a solution in a beaker as part of a laboratory experiment on energy transfer. After a while, she noticed the liq
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The aluminum atom______electrons to form an ion.<br> The ion that is formed is______.
Alchen [17]

The aluminum atom_loses_____electrons to form an ion.

The ion that is formed is_Al³⁺_____.

aluminium has the electronic configuration as 1s² 2s² 2p⁶ 3s² 3p¹

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after losing 3 electrons , the ion formed is given as Al³⁺

hence the correct options to fill in the blanks are lose  and Al³⁺


3 0
3 years ago
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A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi
Masteriza [31]

Answer:

a)S_1=-9.65}\ J/K

b)S_2=71.18\ J/K

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

S_1=-\dfrac{7190}{745}\ J/K

Negative sign because heat is leaving.

S_1=-9.65}\ J/K

b)

The entropy change of reservoir 101 K

S_2=\dfrac{7190}{101}\ J/K

S_2=71.18\ J/K

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0
3 years ago
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