Answer:
Explanation:
Given:
Force, f = 5 N
Velocity, v = 5 m/s
Power, p = energy/time
Energy = mass × acceleration × distance
Poer, p = force × velocity
= 5 × 5
= 25 W.
Note 1 watt = 0.00134 horsepower
But 25 watt,
0.00134 hp/1 watt × 25 watt
= 0.0335 hp.
Answer:
12.5752053801 m/s
![80\times 10^{-3}\ m^3/s](https://tex.z-dn.net/?f=80%5Ctimes%2010%5E%7B-3%7D%5C%20m%5E3%2Fs)
No.
Explanation:
Q = Volume flow rate = ![80\ L/s=80\times 10^{-3}\ m^3/s](https://tex.z-dn.net/?f=80%5C%20L%2Fs%3D80%5Ctimes%2010%5E%7B-3%7D%5C%20m%5E3%2Fs)
d = Diameter of pipe = 9 cm
A = Area = ![\dfrac{\pi}{4}d^2](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpi%7D%7B4%7Dd%5E2)
Volume flow rate is given by
![Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s](https://tex.z-dn.net/?f=Q%3DAv%5C%5C%5CRightarrow%20v%3D%5Cdfrac%7BQ%7D%7BA%7D%5C%5C%5CRightarrow%20v%3D%5Cdfrac%7B80%5Ctimes%2010%5E%7B-3%7D%7D%7B%5Cdfrac%7B%5Cpi%7D%7B4%7D%20%289%5Ctimes%2010%5E%7B-2%7D%29%5E2%7D%5C%5C%5CRightarrow%20v%3D12.5752053801%5C%20m%2Fs)
Velocity of fluid is 12.5752053801 m/s
The volume flow rate in m³/s is ![80\times 10^{-3}\ m^3/s](https://tex.z-dn.net/?f=80%5Ctimes%2010%5E%7B-3%7D%5C%20m%5E3%2Fs)
The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.
Answer:
Yes, a mixture can be made up of just elements and no compounds. The elements never experiance a chemical interaction and stay only in physical contact with each other.
BRAILIEST PLS I NEED LEVEL UP!
Answer:
It is impossible to detect underground water from the surface. Dowsing practitioners refuse to explain their secrets.
Explanation:
Answer:
a)
1.35 kg
b)
2.67 ms⁻¹
Explanation:
a)
= mass of first body = 2.7 kg
= mass of second body = ?
= initial velocity of the first body before collision = ![v](https://tex.z-dn.net/?f=v)
= initial velocity of the second body before collision = 0 m/s
= final velocity of the first body after collision =
using conservation of momentum equation
![m_{1} v_{1i} + m_{2} v_{2i} = m_{1} v_{1f} + m_{2} v_{2f}\\(2.7) v + m_{2} (0) = (2.7) (\frac{v}{3} ) + m_{2} v_{2f}\\(2.7) (\frac{2v}{3} ) = m_{2} v_{2f}\\v_{2f} = \frac{1.8v}{m_{2}}](https://tex.z-dn.net/?f=m_%7B1%7D%20v_%7B1i%7D%20%2B%20m_%7B2%7D%20v_%7B2i%7D%20%3D%20m_%7B1%7D%20v_%7B1f%7D%20%2B%20m_%7B2%7D%20v_%7B2f%7D%5C%5C%282.7%29%20v%20%2B%20m_%7B2%7D%20%280%29%20%3D%20%282.7%29%20%28%5Cfrac%7Bv%7D%7B3%7D%20%29%20%2B%20m_%7B2%7D%20v_%7B2f%7D%5C%5C%282.7%29%20%28%5Cfrac%7B2v%7D%7B3%7D%20%29%20%3D%20m_%7B2%7D%20v_%7B2f%7D%5C%5Cv_%7B2f%7D%20%3D%20%5Cfrac%7B1.8v%7D%7Bm_%7B2%7D%7D)
Using conservation of kinetic energy
![m_{1} v_{1i}^{2}+ m_{2} v_{2i}^{2} = m_{1} v_{1f}^{2} + m_{2} v_{2f}^{2} \\(2.7) v^{2} + m_{2} (0)^{2} = (2.7) (\frac{v}{3} )^{2} + m_{2} (\frac{1.8v}{m_{2}})^{2} \\(2.7) = (0.3) + \frac{3.24}{m_{2}}\\m_{2} = 1.35](https://tex.z-dn.net/?f=m_%7B1%7D%20v_%7B1i%7D%5E%7B2%7D%2B%20m_%7B2%7D%20v_%7B2i%7D%5E%7B2%7D%20%3D%20m_%7B1%7D%20v_%7B1f%7D%5E%7B2%7D%20%2B%20m_%7B2%7D%20v_%7B2f%7D%5E%7B2%7D%20%5C%5C%282.7%29%20v%5E%7B2%7D%20%2B%20m_%7B2%7D%20%280%29%5E%7B2%7D%20%3D%20%282.7%29%20%28%5Cfrac%7Bv%7D%7B3%7D%20%29%5E%7B2%7D%20%2B%20m_%7B2%7D%20%28%5Cfrac%7B1.8v%7D%7Bm_%7B2%7D%7D%29%5E%7B2%7D%20%5C%5C%282.7%29%20%3D%20%280.3%29%20%2B%20%5Cfrac%7B3.24%7D%7Bm_%7B2%7D%7D%5C%5Cm_%7B2%7D%20%3D%201.35)
b)
= mass of first body = 2.7 kg
= mass of second body = 1.35 kg
= initial velocity of the first body before collision = 4 ms⁻¹
= initial velocity of the second body before collision = 0 m/s
Speed of the center of mass of two-body system is given as
ms⁻¹