Just find the density of every metal and select the one with a density of 2.71 g/cm³ . This is:
Metal 1
ρ = m/V
ρ = 22.1 g / 3 cm³
ρ = 7.367 g / cm³
Metal 2
ρ = m/V
ρ = 42 g / 4 cm³
ρ = 10.5 g / cm³
Metal 3
ρ = m/V
ρ = 9.32 g / 5 cm³
ρ = 1.864 g / cm³
Metal 4
ρ = m/V
ρ = 8.13 g / 3 cm³
ρ = 2.71 g / cm³
<h2>R / Metal 4 was selected.</h2>
The sound wave would behave differently in a swimming pool than in his bedroom because sound waves travel faster in more dense mediums; such as water. The wave will travel faster in water, and slower in air.
Reduce friction because friction just makes things harder
Answer:
The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Explanation:
Given:-
- The diameter of the drill bit, d = 98 cm
- The power at which drill works, P = 5.85 hp
- The rotational speed of drill, N = 1900 rpm
Find:-
What Torque And Force Is Applied To The Drill Bit?
Solution:-
- The amount of torque (T) generated at the periphery of the cutting edges of the drilling bit when it is driven at a power of (P) horsepower at some rotational speed (N).
- The relation between these quantities is given:
T = 5252*P / N
T = 5252*5.85 / 1900
T = 16.171 Nm
- The force (F) applied at the periphery of the drill bit cutting edge at a distance of radius from the center of drill bit can be determined from the definition of Torque (T) being a cross product of the Force (F) and a moment arm (r):
T = F*r
Where, r = d / 2
F = 2T / d
F = 2*16.171 / 0.98
F = 33 N
Answer: The torque applied by the drill bit is T = 16.2 Nm and the cutting force of the drill bit is F = 33 N.
Answer:
c. 0.02 C and 4 J
Explanation:
Applying,
Q = CV................ Equation 1
Where Q = Charge, C = Capacitance of the capacitor, V = Voltage.
From the question,
Given: C = 50 μF = 50×10⁻⁶ F, V = 400 V
Substitute these values into equation 1
Q = (50×10⁻⁶)(400)
Q = 0.02 C.
Also Applying
E = CV²/2............. Equation 2
Where E = Energy stored.
Therefore,
E = (50×10⁻⁶ )(400²)/2
E = 4 J
Hence the right option is c. 0.02 C and 4 J
