The force required to pull the two hemispheres is 46622.72N
<h3>Calculation and Parameters</h3>
( Note: 1 millibar=100 N/m2. One atmosphere is 1013 millibar = 1.013×105 N/m2 ]
The contact area between the hemispheres is (pi x 0.400^2) = 0.5024m^2.
Pressure difference = (940 - 12)
= 928 millibars.
(928 x 100)
= 92,800N/m^2.
Therefore, the required force to pull the two hemispheres is
(92800 x 0.5024)
= 46622.72N.
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We have: v i (initial velocity) = 6 m/sv = 1.1 m/sa = - 9.8 m/s²v = v i + a · t1.1 m/s = 6 m/s - 9.8 m/s² t9.8 t = 6 - 1.19.8 t = 4.9t = 4.9 : 9.8t = 0.5 sThen the replacement:x = xi + vi · t + a t² / 2( xi = 0 )x = 6 · 0.5 - 9.8 · 0.25 / 2x = 3 - 1.225Answer:
x = 1.775 m
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Plan: Use Q = m · c · ΔT three times. Hot casting cools ΔT_hot = 500°C -
Tf. Cold water and steel tank heat ΔT_cold = Tf - 25°C. Set Q from hot
casting cooling = Q from cold tank heating.
here
m_cast · c_steel · ΔT_hot = (m_tank · c_steel + m_water · c_water) · ΔT_cold
m_cast · c_steel · (500°C - Tf) = (m_tank · c_steel + m_water · c_water) · (Tf - 25°C)
2.5 kg · 0.50 kJ/(kg K°) · (500°C - Tf) = (5 kg· 0.50 kJ/(kg K°) + 40 kg· 4.18 kJ/(kg K°)) · (Tf - 25°C)
Solve for Tf, remember that K° = C° (i.e. for ΔT's) </span>
Answer:
Resultant horizontal force = 143 N
Explanation:
Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;
Resultant horizontal force = 165 * cos 30
Resultant horizontal force = 142.89
Approximating to a whole number gives;
Resultant horizontal force = 143 N
Explanation:
physical quantity is any physical property that can be qualified that,is, be measured using numbers e.g mass, amount of substance,time and length
