Answer:
joule
Explanation:
Let m = mass of the car and v1 = initial velocity and v2 = final velocity
Given.
Initial velocity = 100 km/h
final velocity = 50 km/h
What is work done in the car to slow it from 100km/h to 50km/h?
The work done in the car to slow it from v1 to v2.
w=Δk
joule.
Therefore, the work done is joule
Answer:
2.846m
Explanation:
The diver is performing projectile motion.
To find x(final), we are going to use the equation x(final) = v(initial)*t + x(initial)
x(initial) = 0
x(final) = ?
v(initial) = 2.3 m/s
we don't know t
To find t we will use y(final) = 1/2*(-9.8)*t^2 + v(initial in the y dir.)*t + y(initial)
- 9.8 in the acceleration in the y dir.
y(final) = 0
y(initial) = 7.5
v(initial in the y dir.) = 0
If we solve for t we get: t = 1.237s
Now we have all the components to solve for x(final) in x(final) = v(initial)*t + x(initial)
x(final) = 2.3*1.237 + 0
x(final) = 2.846m
Answer:
When the obstacle is fixed, the law of action and reaction, makes the reflected wave is inverted.
When the obstacle is mobile, he mobile point, it moves in the direction of the wave, therefore there is no inversion of it.
Explanation:
Waves when they reach an obstacle behave like a shock, therefore if we use the conservation of momentum the wave must reverse its speed, this explains that the speed changes sign, the wave is reflected.
When the obstacle is fixed, the wave when it reaches the obstacle exerts a force on the point, by the law of action and reaction the point exerts on the wave a force of equal magnitude but in the opposite direction, this reaction force which makes the reflected wave is inverted.
When the obstacle is mobile, this is without friction, when the wave arrives it exerts a force on the mobile point, it moves in the direction of the wave, reaching the maximum amplitude of the incident wave, when it is reflected the point begins to go down along with the wave, therefore there is no inversion of it.
Answer:
Net Force Formula: F1+F2+F3...+FN
*(MAKE SURE YOU MAKE NOTE OF THE NEGATIVE FORCES AND SUBTRACT THEM!)*
Explanation:
Top left: 17N+25N+25N-42N= 25N
Top right: 65N+200N-65N-150N-200N= 150N
Bottom left: 189N+123N+284N-96N-188N-312N= 0N
Bottom right: 34N+77N-12N-34N= 65N
I hope this helps! :))
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)
