A scientist can assess whether a pure niobium sample is responsible for contaminating the lab with radioactivity by testing the sample. By testing the niobium sample, a scientist can determine whether it has any other element.
Answer:
the answer is force . force is applied as a push or pull
Answer:
8977.7 kg/m^3
Explanation:
Volume of water displaced = 55 cm^3 = 55 x 10^-6 m^3
Reading of balance when block is immersed in water = 4.3 N
According to the Archimedes principle, when a body is immersed n a liquid partly or wholly, then there is a loss in the weight of body which is called upthrust or buoyant force. this buoyant force is equal to the weight of liquid displaced by the body.
Buoyant force = weight of the water displaced by the block
Buoyant force = Volume of water displaced x density of water x g
= 55 x 10^-6 x 1000 x .8 = 0.539 N
True weight of the body = Weight of body in water + buoyant force
m g = 4.3 + 0.539 = 4.839
m = 0.4937 kg
Density of block = mass of block / volume of block
= 
Density of block = 8977.7 kg/m^3
The potential difference across the parallel plate capacitor is 2.26 millivolts
<h3>Capacitance of a parallel plate capacitor</h3>
The capacitance of the parallel plate capacitor is given by C = ε₀A/d where
- ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
- A = area of plates and
- d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.
<h3>Charge on plates</h3>
Also, the surface charge on the capacitor Q = σA where
- σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
- a = area of plates.
<h3>
The potential difference across the parallel plate capacitor</h3>
The potential difference across the parallel plate capacitor is V = Q/C
= σA ÷ ε₀A/d
= σd/ε₀
Substituting the values of the variables into the equation, we have
V = σd/ε₀
V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m
V = 20.0 C/m × 10⁻³/8.854 F/m
V = 2.26 × 10⁻³ Volts
V = 2.26 millivolts
So, the potential difference across the parallel plate capacitor is 2.26 millivolts
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