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juin [17]
3 years ago
14

a television of mass 8kg sits on a table the coefficient of static friction between the table and television is .48 what is the

minimum applied force that will cause the television to slide
Physics
1 answer:
Mama L [17]3 years ago
3 0
Th normal reaction between the television and the table is
N = (8 kg)(9.8 m/s²) = 78.4 N

The static coefficient of friction is μ = 0.48.

When the television is about to slide on the table, the applied force should overcome the force due to static friction.

The applied force should be at least
F = μN
   = 0.48*78.4
   = 37.632 N

Answer: 37.6 N (nearest tenth)

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Explain the differences in structure and use for life between oxygen gas in the atmosphere and ozone.
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Answer:

Most of the oxygen in the atmosphere is in the form of O2, which means it is made up of molecules containing two oxygen atoms. Ozone, however is O3, which means it is made up of molecules containing three oxygen atoms. O2 is what we breath, and what plants release from photosynthesis. Ozone occurs naturally high in the stratosphere, where it absorbs ultraviolet light, protecting us here on the surface from skin cancer. Ozone can also occur closer to the surface of the earth as a pollutant. It is formed from reactions with O2 and chemicals emitted from factories and cars. It comes in the form of smog.

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3 years ago
What ocean depth would the volume of an aluminium sphere be reduced by 0.10%
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Answer:

6400 m

Explanation:

You need to use the bulk modulus, K:

K = ρ dP/dρ

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Since ρ is changing by very little, we can say:

K ≈ ρ ΔP/Δρ

Therefore, solving for ΔP:

ΔP = K Δρ / ρ

We can calculate K from Young's modulus (E) and Poisson's ratio (ν):

K = E / (3 (1 - 2ν))

Substituting:

ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)

Before compression:

ρ = m / V

After compression:

ρ+Δρ = m / (V - 0.001 V)

ρ+Δρ = m / (0.999 V)

ρ+Δρ = ρ / 0.999

1 + (Δρ/ρ) = 1 / 0.999

Δρ/ρ = (1 / 0.999) - 1

Δρ/ρ = 0.001 / 0.999

Given:

E = 69 GPa = 69×10⁹ Pa

ν = 0.32

ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)

ΔP = 64.0×10⁶ Pa

If we assume seawater density is constant at 1027 kg/m³, then:

ρgh = P

(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa

h = 6350 m

Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.

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