Answer:
Force is measured in "Newton's" - kg-m/s^2
F = M a = kg * m/s^2
This in the kgm-m systen - in the gram-cm system you have
F = dynes = gram-cm/s^2
Since 1 gm = .001 kg and 1 cm = .01 m
1 dyne = 1.0^-5 newtons
The ball spins once around the circle in 0.20 s, meaning it travels a distance equal to its circumference in that time, giving it a linear speed of
<em>v</em> = (2<em>π</em> (0.5 m)) / (0.20 s) = 5<em>π</em> m/s ≈ 15.708 m/s
Use this compute the magnitude of the centripetal acceleration <em>a</em> :
<em>a</em> = <em>v</em>²/ (0.5 m) = 50<em>π</em>² m/s² ≈ 493.48 m/s²
Use Newton's second law to compute the mangitude of the tension <em>F</em> in the string:
<em>F</em> = (0.05 kg) <em>a</em> = 5/2 <em>π</em>² N ≈ 24.674 N ≈ 25 N
Answer:
E =9525.6 J
Explanation:
Given that
Current ,I = 4.9 A
time ,t = 5.4 min
Voltage difference ,ΔV = 6 V
energy delivered by battery in time t is given as
E=V I t
where V is voltage, I is the current and t is the time.
now putting the values in above equation
E= 6 x 4.9 x 5.4 x 60 J ( 1 min = 60 s , 5.4 min = 5.4 x 60 s)
E =9525.6 J
Therefore the energy reduced will be 9525.6 J
Answer:
![P = 3.55 \times 10^5 Pa](https://tex.z-dn.net/?f=P%20%3D%203.55%20%5Ctimes%2010%5E5%20Pa)
Explanation:
As we know that water from the fountain will raise to maximum height
![H = 26.0 m](https://tex.z-dn.net/?f=H%20%3D%2026.0%20m)
now by energy conservation we can say that initial speed of the water just after it moves out will be
![\frac{1}{2}mv^2 = mgH](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%3D%20mgH)
![v = \sqrt{2gH}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gH%7D)
![v = \sqrt{2(9.81)(26)}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2%289.81%29%2826%29%7D)
![v = 22.6 m/s](https://tex.z-dn.net/?f=v%20%3D%2022.6%20m%2Fs)
Now we can use Bernuolli's theorem to find the initial pressure inside the pipe
![P = P_0 + \frac{1}{2}\rho v^2](https://tex.z-dn.net/?f=P%20%3D%20P_0%20%2B%20%5Cfrac%7B1%7D%7B2%7D%5Crho%20v%5E2)
![P = 10^5 + \frac{1}{2}(1000)(22.6^2)](https://tex.z-dn.net/?f=P%20%3D%2010%5E5%20%2B%20%5Cfrac%7B1%7D%7B2%7D%281000%29%2822.6%5E2%29)
![P = 3.55 \times 10^5 Pa](https://tex.z-dn.net/?f=P%20%3D%203.55%20%5Ctimes%2010%5E5%20Pa)
Answer:
that one i know only pe not that sorry again