The kinetic energy of an object is given by:
![K= \frac{1}{2}mv^2](https://tex.z-dn.net/?f=K%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20)
where m is the mass of the object and v its velocity.
The car in this problem has a mass of m=600 kg and a velocity of v=10 m/s, therefore if we put these numbers into the equation, we find the kinetic energy of the car:
Answer:
The driver can avoid the child because at the time she brake 1.6s the car just move 20.928 meters so is far to the 65 meters where the kind is
Distance= 20.928 m
Explanation:
![V_{o} = 17 \frac{m}{s} \\a= -4.9 \frac{m}{s^{2} }\\ t= 1.6 s](https://tex.z-dn.net/?f=V_%7Bo%7D%20%3D%2017%20%5Cfrac%7Bm%7D%7Bs%7D%20%5C%5Ca%3D%20-4.9%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%5C%5C%20t%3D%201.6%20s)
Distance the kid is 65m so the car have to be less in the time she is braking
So to calculated the distance while she is braking
![X_{f} = X_{o} +V_{o}*t + \frac{1}{2} * a *t^{2}\\ X_{f} = 0m +17 \frac{m}{s} *1.6 s + \frac{1}{2} * 4.9 \frac{m}{s^{2} } *(1.6s)^{2}\\X_{f} = 27.2 m+ (-6.272 m)\\X_{f} = 20.928 m](https://tex.z-dn.net/?f=X_%7Bf%7D%20%3D%20X_%7Bo%7D%20%2BV_%7Bo%7D%2At%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%20a%20%2At%5E%7B2%7D%5C%5C%20X_%7Bf%7D%20%3D%200m%20%2B17%20%5Cfrac%7Bm%7D%7Bs%7D%20%2A1.6%20s%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%2A%204.9%20%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%20%7D%20%2A%281.6s%29%5E%7B2%7D%5C%5CX_%7Bf%7D%20%3D%2027.2%20m%2B%20%28-6.272%20m%29%5C%5CX_%7Bf%7D%20%3D%2020.928%20m)
The distance of the car is less than the distance of the kid in his bike
So she didn't hit him
Answer:
A = 0.22 m
Explanation:
The spring with the block and the pebble forms an oscillatory system, this system is described by the expression
x = A cos (wt + φ)
w = √ (k / m).
The data they give us is the amplitude of the movement (A = 10 cm), the oscillation mass is equal to the block mass plus the mass of the pebble
m = m + M
m = 0.031 + 0.108
m = 0. 139 kg
To find the spring constant let's use Hooke's law
F = k X
The force is the weight of the pebble and the additional elongation is x = 4.9 cm, let's calculate
k = F / x
k = mg / x
k = 0.031 9.8 / 0.049
k = 6.2 N / m
Let's look for angular velocity
w = √ (6.2 / 0.139)
w = 6,670 rad / s
Let's write the oscillation equation with this data
x = 0.10 cos (6,670 t)
For the pebble to remain in contact with the block, the acceleration of the spring system plus block with pebble must be less than the acceleration of gravity in the descending oscillation
Let's look for system acceleration
a = d²x / dt²
dx / dt = - A w sin (wt + Ф)
d²x / dt² = - A w² cos (wt+Ф)
To find the maximum value cos (wt) = ±1
g = A w²
A = g / w²
A = 9.8 / 6.67²
A = 0.22 m
When the amplitude of the oscillation exceeds this value the pebble is delayed with respect to the block
We don't know that at all. The 3rd law says that the REaction is opposite and EQUAL to the action. We don't know where that "twice as much" comes from.
Answer:
A) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent
Explanation:
When it is found that you are out of air while under water, first of all don't panic, look for your buddy. If you are unable to do that so, then you need to make an emergency ascent. First try to make a <u>Controlled Emergency Swimming Ascent</u> (CESA). This ascent remains under control and is performed at a safe ascent rate. As you ascend the air in your lungs will expand with decreasing ambient pressure. To avoid an over pressurization injury, always exhale a continuous string of bubbles while going up.
If you are not sure you will make to the surface that leading to inhale the only option is to turn the CESA into a Buoyant Emergency Ascent. To be ready locate your weight system as you ascend. As an addition remove the weight from one of weight pockets and hold it away from your body in preparation of dropping if necessary. Dropping the weight will give you an upward buoyant force which is an <u>uncontrolled buoyant emergency ascent</u> and <u><em>should be performed only as the last option</em></u>.
So, according to this, first, always have a look at your SPG. Then, if you are out of air, look for your buddy, if not found then make CESA and the last option will be buoyant emergency ascent.