Answer:
a
Explanation:
because it has more energy
We know that impulse is simply the product of Force and time:
Impulse = Force * time
Since Force has a unit of Newton or kg m/s^2 and time is in
seconds, therefore impulse can have units as:
N s
or
<span>kg m/s</span>
Explanation:
If you cannot visualize it, just assume that the distance from station A to B is 420km. Each half is 210km.
When the car travels from A to B, it takes 420/30 = 14 hours.
When the car travels from B to the halfway point, it takes 210/30 = 7 hours.
When the car travels from the halfway point to A, it takes 210/70 = 3 hours.
Total time taken = 14 + 7 + 3 = 24 hours.
Total distance = 420km * 2 = 840km.
Hence, the average speed of the car is 840/24 = 35km/h.
Explanation:
formula: <u>Mass</u>
Density x volume
2a) m=10kg v=0.3m³
10÷0.3=33.3 kg/m
2b) m = 160 kg V=0.1m³
160÷0.1=1600 kg/m
2c) m = 220 kg V = 0.02m³
220÷0.02=11000 kg/m
A wooden post has a volume of 0.025m³ and a mass of 20kg. Calculate its density in kg/m.
density = volume ÷ mass
20÷ 0.025=800 kg/m
Challenge: A rectangular concrete slab is 0.80m long, 0.60 m wide and 0.04m thick. Calculate its volume in m³.
Formula : Length x width x height = Volume
0.80 x 0.60 x 0.04 = 0.0192m³
B) The mass of the concrete slab is 180 kg. Calculate its density in kg/m.
density = volume ÷ mass
180 ÷ 0.0192 = 9375 kg/m
Given data:
* The mass of the baseball is 0.31 kg.
* The length of the string is 0.51 m.
* The maximum tension in the string is 7.5 N.
Solution:
The centripetal force acting on the ball at the top of the loop is,
![\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T%2Bmg%3D%5Cfrac%7Bmv%5E2%7D%7BL%7D_%7B%7D%20%5C%5C%20v%5E2%3D%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%20%5C%5C%20v%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T%2Bmg%29%7D%7Bm%7D%7D%20%5Cend%7Bgathered%7D)
For the maximum velocity of the ball at the top of the vertical circular motion,
![v_{\max }=\sqrt[]{\frac{L(T_{\max }+mg)}{m}}](https://tex.z-dn.net/?f=v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7BL%28T_%7B%5Cmax%20%7D%2Bmg%29%7D%7Bm%7D%7D)
where g is the acceleration due to gravity,
Substituting the known values,
![\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%287.5_%7B%7D%2B0.31%5Ctimes9.8%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B%5Cfrac%7B0.51%2810.538%29%7D%7B0.31%7D%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D%5Csqrt%5B%5D%7B17.34%7D%20%5C%5C%20v_%7B%5Cmax%20%7D%3D4.16%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.
