Answer:
The time interval is ![t = 3 \ s](https://tex.z-dn.net/?f=t%20%3D%20%203%20%5C%20s)
Explanation:
From the question we are told that
The angular acceleration is ![\alpha = 4.0 \ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%204.0%20%5C%20rad%2Fs%5E2)
The time taken is ![t = 4.0 \ s](https://tex.z-dn.net/?f=t%20%20%3D%20%204.0%20%5C%20s)
The angular displacement is ![\theta = 80 \ radians](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%2080%20%5C%20radians)
The angular displacement can be represented by the second equation of motion as shown below
![\theta = w_i t + \frac{1}{2} \alpha t^2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%20w_i%20t%20%20%2B%20%20%5Cfrac%7B1%7D%7B2%7D%20%5Calpha%20t%5E2)
where
is the initial velocity at the start of the 4 second interval
So substituting values
![80 = w_i * 4 + 0.5 * 4.0 * (4^2)](https://tex.z-dn.net/?f=80%20%3D%20%20w_i%20%2A%20%204%20%2B%200.5%20%2A%20%204.0%20%2A%20%20%284%5E2%29)
=> ![w_i = 12 \ rad/s](https://tex.z-dn.net/?f=w_i%20%20%3D%20%2012%20%5C%20rad%2Fs)
Now considering this motion starting from the start point (that is rest ) we have
![w__{4.0 }} = w__{0}} + \alpha * t](https://tex.z-dn.net/?f=w__%7B4.0%20%7D%7D%20%3D%20%20w__%7B0%7D%7D%20%2B%20%20%5Calpha%20%20%2A%20t)
Where
is the angular velocity at rest which is zero and
is the angular velocity after 4.0 second which is calculated as 12 rad/s s
![12 = 0 + 4 t](https://tex.z-dn.net/?f=12%20%3D%20%200%20%20%2B%204%20t)
=> ![t = 3 \ s](https://tex.z-dn.net/?f=t%20%3D%20%203%20%5C%20s)
Answer:
C)It would require more energy to change solid water into liquid water because there are more molecules in this larger piece of ice.
Answer: C
Explanation:
A bird flying in the air is 3D motion as the bird move up, down, right, left all the possible direction. A leaf falling from a tree is also a 3D motion because a leaf never fall in absolute vertical manner. A lady bug crawling on a soccer ball also 3D. If lady bug crawling on a plane like floor then it is 2D but now lady bug crawling on soccer ball which is in spherical shape so it is also 3D motion. Only train travelling along a track is 2D because train cannot move up and down and the track of train is in a plane. So it is 2D motion.
Answer:
a
![H =212.6 \ J](https://tex.z-dn.net/?f=H%20%20%3D212.6%20%5C%20%20J)
b
![v = 7.647 \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%207.647%20%20%5C%20%20m%2Fs)
Explanation:
From the question we are told that
The child's weight is ![W_c = 287 \ N](https://tex.z-dn.net/?f=W_c%20%20%3D%20%20287%20%5C%20N)
The length of the sliding surface of the playground is ![L = 7.20 \ m](https://tex.z-dn.net/?f=L%20%3D%20%207.20%20%5C%20%20m)
The coefficient of friction is ![\mu = 0.120](https://tex.z-dn.net/?f=%5Cmu%20%3D%20%200.120)
The angle is ![\theta = 31.0 ^o](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2031.0%20%5Eo)
The initial speed is ![u = 0.559 \ m/s](https://tex.z-dn.net/?f=u%20%3D%20%200.559%20%5C%20%20m%2Fs)
Generally the normal force acting on the child is mathematically represented as
=> ![N = mg * cos \theta](https://tex.z-dn.net/?f=N%20%20%3D%20%20mg%20%20%2A%20%20cos%20%5Ctheta)
Note ![m * g = W_c](https://tex.z-dn.net/?f=m%20%2A%20%20g%20%20%3D%20%20W_c)
Generally the frictional force between the slide and the child is
![F_f = \mu * mg * cos \theta](https://tex.z-dn.net/?f=F_f%20%20%3D%20%20%5Cmu%20%2A%20%20mg%20%20%2A%20%20cos%20%5Ctheta)
Generally the resultant force acting on the child due to her weight and the frictional force is mathematically represented as
![F =m* g sin(\theta) - F_f](https://tex.z-dn.net/?f=F%20%3Dm%2A%20g%20sin%28%5Ctheta%29%20-%20F_f)
Here F is the resultant force and it is represented as ![F = ma](https://tex.z-dn.net/?f=F%20%3D%20%20ma)
=> ![ma = m* g sin(31.0) - \mu * mg * cos (31.0)](https://tex.z-dn.net/?f=ma%20%3D%20%20%20m%2A%20g%20sin%2831.0%29%20%20-%20%5Cmu%20%2A%20%20mg%20%20%2A%20%20cos%20%2831.0%29)
=> ![a = g sin(31.0)- \mu * g * cos (31.0)](https://tex.z-dn.net/?f=a%20%3D%20%20g%20sin%2831.0%29-%20%20%5Cmu%20%2A%20%20g%20%20%2A%20%20cos%20%2831.0%29)
=> ![a = 9.8 * sin(31.0) - 0.120 * 9.8 * cos (31.0)](https://tex.z-dn.net/?f=a%20%3D%20%20%20%209.8%20%2A%20%20sin%2831.0%29%20-%200.120%20%2A%20%209.8%20%20%2A%20%20cos%20%2831.0%29)
=>![a = 4.039 \ m/s^2](https://tex.z-dn.net/?f=a%20%3D%20%204.039%20%5C%20m%2Fs%5E2)
So
![F_f = 0.120 * 287 * cos (31.0)](https://tex.z-dn.net/?f=F_f%20%20%3D%20%200.120%20%20%2A%20287%20%20%2A%20%20cos%20%2831.0%29)
=> ![F_f = 29.52 \ N](https://tex.z-dn.net/?f=F_f%20%20%3D%2029.52%20%5C%20%20N)
Generally the heat energy generated by the frictional force which equivalent tot the workdone by the frictional force is mathematically represented as
![H = F_f * L](https://tex.z-dn.net/?f=H%20%20%3D%20%20F_f%20%20%2A%20L)
=> ![H = 29.52 * 7.2](https://tex.z-dn.net/?f=H%20%20%3D%2029.52%20%2A%20%207.2)
=> ![H =212.6 \ J](https://tex.z-dn.net/?f=H%20%20%3D212.6%20%5C%20%20J)
Generally from kinematic equation we have that
![v^2 = u^2 + 2as](https://tex.z-dn.net/?f=v%5E2%20%20%3D%20%20u%5E2%20%20%2B%20%202as)
=> ![v^2 = 0.559^2 + 2 * 4.039 * 7.2](https://tex.z-dn.net/?f=v%5E2%20%20%3D%20%200.559%5E2%20%20%2B%20%202%20%2A%204.039%20%2A%207.2)
=> ![v = \sqrt{0.559^2 + 2 * 4.039 * 7.2}](https://tex.z-dn.net/?f=v%20%20%3D%20%20%5Csqrt%7B0.559%5E2%20%20%2B%20%202%20%2A%204.039%20%2A%207.2%7D)
=> ![v = 7.647 \ m/s](https://tex.z-dn.net/?f=v%20%20%3D%20%207.647%20%20%5C%20%20m%2Fs)