KINEMATICS
Uniform or constant motion in a straight line (rectilinear). Speed or velocity constant and/or acceleration constant. If motion is up and down and/or has an up and down component then acceleration omn earth will be g. g is about 10m/s/s.
speed = distance/time
velocity = displacement/time
s=distance ... u=initial speed ... v = final speed ... a = acceleration ... t = time
v=u+at
v^2=u^2+2as
s=ut+1/2at^2
<em>Answers:</em>
<em>1. The hot soup will loose the heat and the ice water will gain the heat</em>
- If two jars are insulated inside the insulated box, the heat may not be transferred to outside of the box.
- According to II law of thermodynamics, Heat always flow from high temperature body to low temperature body, with out aid of external energy.
<em>So, from two points, it is concluded that The hot soup will lose heat and the ice water will gain the heat until they reach the thermodynamic equilibrium.</em>
<em>2. The particles in gases are farther apart and move faster </em>
- Particles in the gases are loosely packed (greater distance between particles compared to solids and liquids) and particles collide less often.
<em>Therefore conduction is weak in gases compared to solids and liquids.</em>
<em>3. Heat and milk by conduction; popping popcorn by radiation.</em>
- The heat can transfer from pot to the milk by conduction because they are in contact at boundaries, similarly the pot and the stove are in contact <em>so the conduction transfers heat from pot to the milk. </em>
- In microwave oven there is no direct contact (<em>no conduction</em>) of heat and popcorn, also there is no molecular momentum transfer <em>(means of no convection).</em>
<em>So obviously the heat transfer by radiation occurs in a microwave oven.</em>
Answer:
The answer is "effective stress at point B is 7382 ksi
"
Explanation:
Calculating the value of Compressive Axial Stress:
![\to \sigma y =\frac{F}{A} = \frac{4 F}{( p d ^2 )} = \frac{(4 x ( - 40000 \ lbf))}{[ p \times (1 \ in)^2 ]} = - 50.9 \ ksi \\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%20y%20%20%3D%5Cfrac%7BF%7D%7BA%7D%20%3D%20%5Cfrac%7B4%20F%7D%7B%28%20p%20d%20%5E2%20%29%7D%20%3D%20%5Cfrac%7B%284%20x%20%28%20-%2040000%20%5C%20lbf%29%29%7D%7B%5B%20p%20%5Ctimes%20%281%20%5C%20in%29%5E2%20%5D%7D%20%3D%20-%2050.9%20%5C%20ksi%20%5C%5C)
Calculating Shear Transverse:
![\to \sigma' =[ s y^2 +3( t \times y^2 + t yz^2 )] \times \frac{1}{2}\\\\](https://tex.z-dn.net/?f=%5Cto%20%5Csigma%27%20%3D%5B%20s%20y%5E2%20%2B3%28%20t%20%5Ctimes%20y%5E2%20%2B%20t%20yz%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C)
![= [ (-50.9)^2 +3((63.7)^2 +(0.17)^2 )] \times \frac{1}{2}\\\\=[2590.81+ 3(4057.69)+0.0289]\times \frac{1}{2}\\\\=[2590.81+12,173.07+0.0289] \times \frac{1}{2}\\\\=14763.9089\times \frac{1}{2}\\\\ = 7381.95445 \ ksi\\\\ = 7382 \ ksi](https://tex.z-dn.net/?f=%3D%20%5B%20%28-50.9%29%5E2%20%2B3%28%2863.7%29%5E2%20%2B%280.17%29%5E2%20%29%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B%203%284057.69%29%2B0.0289%5D%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D%5B2590.81%2B12%2C173.07%2B0.0289%5D%20%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%3D14763.9089%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%5C%5C%5C%5C%20%3D%207381.95445%20%5C%20ksi%5C%5C%5C%5C%20%3D%207382%20%5C%20ksi)
Answer:
Yes, the car driver is exceeding the given limit.
Explanation:
<u>Given:</u>
- Speed of the car, v = 38.0 m/s.
- Speed limit of the highway,
<h2><u>
Converting the speed limit from mi/h to m/s:</u></h2>
We know,
1 mi = 1.60934 km.
1 km = 1000 m.
Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.
1 hour = 60 minutes.
1 minute = 60 seconds.
Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.
Using these values,
Therefore,
Clearly,
which means, the car driver is exceeding the given speed limit.
