Given:
The mass of the halfback is m = 107 kg
The speed of the halfback is v = 8 m/s
To find the momentum.
Explanation:
The momentum of the halfback is
Thus, the momentum of the halfback is 856 kg m/s
Answer:
Note that there is little variation among the transition metals. Electronegativities generally decrease from top to bottom within a group due to the larger atomic size. Of the main group elements, fluorine has the highest electronegativity (EN = 4.0) and cesium the lowest (EN = 0.79).
Angle of incidence is 36° and so is the reflection. Both angles are equal.
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Answer:
The specific heat for the metal is 0.466 J/g°C.
Explanation:
Given,
Q = 1120 Joules
mass = 12 grams
T₁ = 100°C
T₂ = 300°C
The specific heat for the metal can be calculated by using the formula
Q = (mass) (ΔT) (Cp)
ΔT = T₂ - T₁ = 300°C - 100°C = 200°C
Substituting values,
1120 = (12)(200)(Cp)
Cp = 0.466 J/g°C.
Therefore, specific heat of the metal is 0.466 J/g°C.
