Answer:
12m/s2
Explanation:
:D (just a smiley, not answer d)
Answer:
Explanation:
Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.
Therefore,
u = 10 m/s, initial upward velocity.
H = - 20 m, position of the ground.
g = 9.8 m/s², acceleration due to gravity.
Part (a)
When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore
u² - 2gh = 0
h = u²/(2g) = 10²/(2*9.8) = 5.102 m
At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.
Answer: 25.1 m above ground
Part (b)
Let v = the velocity when the frog hits the ground. Then
v² = u² - 2gH
v² = 10² - 2*9.8*(-20) = 492
v = 22.18 m/s
Answer: The frog hits the ground with a velocity of 22.2 m/s
Answer:
3.964 s
Explanation:
Metric unit conversion:
1 miles = 1.6 km = 1600 m.
1 hour = 60 minutes = 3600 seconds
75 mph = 75 * 1600 / 3600 = 33.3 m/s
22.5 mph = 22.5 * 1600/3600 = 10 m/s
Let g = 9.81 m/s2
Friction is the product of coefficient and normal force, which equals to the gravity
![F_f = \mu N = \mu mg](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20N%20%3D%20%5Cmu%20mg)
The deceleration caused by friction is friction divided by mass according to Newton 2nd law.
![a = F_f / m = \mu mg / m = \mu g = 0.6 *9.81 = 5.886 m/s^2](https://tex.z-dn.net/?f=a%20%3D%20F_f%20%2F%20m%20%3D%20%5Cmu%20mg%20%2F%20m%20%3D%20%5Cmu%20g%20%3D%200.6%20%2A9.81%20%3D%205.886%20m%2Fs%5E2)
So the time required to decelerate from 33.3 m/s to 10 m/s so the wheels don't slide, with the rate of 5.886 m/s2 is
![t = \frac{\Delta v}{a} = \frac{33.3 - 10}{5.886} = 3.964 s](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5CDelta%20v%7D%7Ba%7D%20%3D%20%5Cfrac%7B33.3%20-%2010%7D%7B5.886%7D%20%3D%203.964%20s)