Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
1. By predicting the path of hurricanes, it is possible to warn people in its prospective path in advance. This way lives can be saved. In addition economic cost. associated with its destruction, can be minimized by bracing in advance for the storm along its path.
2. When the sea temperatures increase, there is increased vaporization of water into the atmosphere. This causes a transfer of heat into the atmosphere from the hydrosphere. This causes a drop of pressure in the local region that is one ingredient associated with the beginning of a hurricane.
3. The more the humidity, it means the more the heat transfer to the atmosphere through condensation. This reduces the pressure proportionally with the amount of humidity. This low pressure is an ingredient for hurricanes. In addition, the higher the humidity the larger the clouds and the larger the thunderstorms and precipitation associated with the hurricane
4. Winds drive hurricanes and determines their vicious force. In the formation of the hurricane, low winds are required to allow the formation of the funnel at the front betwen the cold and warm air masses. Winds blowing in congruence with the hurricane after then make it stronger while those that blow against it weaken it.
5. The answer is; YES, I would expect a hurricane. This is because at 12 0 N and 45 0 W are the open oceans of the Pacific. The average sea temperaturs in summer is between 15 0 and 20 0. At 27.5 0, this increases the humidity at the local region and therefore the tranfer of heat to the atmsphere. The light breeze gives the stability required for the air convection currents to occur forming a funnel front between the air masses.
At the initial state: v1 = vf = 0.001053 m
3
/kg, h1 = hf = 467.11 kJ/kg, and s1 = sf = 1.4336 kJ/kgK.
The mass of the water is: m = V/v1 = 0.005/0.001053 = 4.7483 kg.
To find the final state, we will use the First Law:
Q12 = m(h2 - h1) for closed system undergoing a constant pressure process.
h2 = 1Q2/m + h1 = 2200/4.7483 + 467.11 = 930.43 kJ/kg.
At P2 = P1 = 150 kPa, this is a saturated mixture.
hf = 467.11 kJ/kg, hfg = 2226.5 kJ/kg, sf = 1.4336 kJ/kgK, and sfg = 5.7897 kJ/kgK
s2 = sf + sfg (h2 – hf )/hfg = 1.4336 + 5.7897(930.43 – 467.11)/2226.5 = 2.6384 kJ/kgK.
The entropy change of water is:
Delta Ssys= m(s2 – s1) = 4.7483(2.6384 – 1.4336) = 5.72 kJ/K.
Answer:
a. 125 kJ
Explanation:
Her total energy is the same as the potential energy she had at the top of the hill:
PE = mgh
= (52 kg)(9.8 m/s^2)(245 m) = 124,852 J
≈ 125 kJ . . . . matches choice A
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After skiing down 112 m, some of her initial energy is converted to kinetic energy, and some remains as potential energy. We assume the ski slope is essentially frictionless, and air resistance is negligible.
