Answer:
F = 1.047 10⁻² N
Explanation:
Let's use kinematics to find the angular acceleration
w = w₀ + α t
as for rest w₀ = 0
w = α t
α = w / t
let's reduce the magnitudes to the SI system
w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s
m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg
r = 10.0 cm (1 m / 100 cm) = 0.100 m
let's calculate
α = 104.72 / 1
α = 104.72 rad / s²
angular and linear variables are related
a = α r
a = 104.72 0.100
a = 10.47 m / s²
finally we substitute in Newton's second law
F = 1 10⁻³ 10.47
F = 1.047 10⁻² N
Answer:
Momentum (P)= 3.15Kg.m/s
Explanation:
Because momentum by definition is the product of mass and velocity, therefore, it's calculated by the formula;
P=mv,,,
where 'P' Is the momentum
'm' is mass in Kilograms, note that mass should be in Kilograms, therefore the 450grams should be converted to kilograms by dividing it with a 1000, i.e (450g×1kg/1000g)= 0.45kg, and now you can substitute in the formula because you have everything you need to find momentum in the right si units.
so P=mv
P=0.45Kg×7m/s
P=3.15Kg.m/s
Answer:
3.7 km
62.7175453187 seconds
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
g = Acceleration due to gravity = 9.81 m/s²
![s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 15\times 10^2\\\Rightarrow s=750\ m](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20s%3D0%5Ctimes%20t%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%2015%5Ctimes%2010%5E2%5C%5C%5CRightarrow%20s%3D750%5C%20m)
![v=u+at\\\Rightarrow v=0+15\times 10\\\Rightarrow v=150\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3D0%2B15%5Ctimes%2010%5C%5C%5CRightarrow%20v%3D150%5C%20m%2Fs)
![s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=150\times 5+\dfrac{1}{2}\times 10\times 5^2\\\Rightarrow s=875\ m](https://tex.z-dn.net/?f=s%3Dut%2B%5Cdfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20s%3D150%5Ctimes%205%2B%5Cdfrac%7B1%7D%7B2%7D%5Ctimes%2010%5Ctimes%205%5E2%5C%5C%5CRightarrow%20s%3D875%5C%20m)
![v=u+at\\\Rightarrow v=150+10\times 5\\\Rightarrow v=200\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3D150%2B10%5Ctimes%205%5C%5C%5CRightarrow%20v%3D200%5C%20m%2Fs)
![v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-200^2}{2\times -9.81}\\\Rightarrow s=2038.73598369\ m](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as%5C%5C%5CRightarrow%20s%3D%5Cdfrac%7Bv%5E2-u%5E2%7D%7B2a%7D%5C%5C%5CRightarrow%20s%3D%5Cdfrac%7B0%5E2-200%5E2%7D%7B2%5Ctimes%20-9.81%7D%5C%5C%5CRightarrow%20s%3D2038.73598369%5C%20m)
The maximum altitude of the rocket is 750+875+2038.73598369 = 3663.73598369 m = 3.7 km
![s=ut+\frac{1}{2}at^2\\\Rightarrow 3663.73598369=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{3663.73598369\times 2}{9.81}}\\\Rightarrow t=27.3301854818\ s](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%203663.73598369%3D0t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209.81%5Ctimes%20t%5E2%5C%5C%5CRightarrow%20t%3D%5Csqrt%7B%5Cfrac%7B3663.73598369%5Ctimes%202%7D%7B9.81%7D%7D%5C%5C%5CRightarrow%20t%3D27.3301854818%5C%20s)
Time taken to reach the ground from max height 27.3301854818 seconds
![v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-200}{-9.81}\\\Rightarrow t=20.3873598369\ s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7Bv-u%7D%7Ba%7D%5C%5C%5CRightarrow%20t%3D%5Cdfrac%7B0-200%7D%7B-9.81%7D%5C%5C%5CRightarrow%20t%3D20.3873598369%5C%20s)
Time to reach max height during free fall after stage 2 ends is 20.3873598369 s
Total time of flight is 10+5+20.3873598369+27.3301854818 = 62.7175453187 seconds
Answer
given,
mass of the shell = 87 g = 0.087 Kg
speed of the muzzle = 853 m/s
mass of the helicopter = 4410 kg
A burst of 176 shell fired in 2.93 s
resulting average force = ?
momentum of the shell = m v
= 0.087 x 853
= 74.21 kgm/s
momentum of 176 shell is = 176 p
= 176 x 74.21
= 13060.96
momentum of helicopter = - 13060.96 kgm/s
amount of speed reduce a = ![\dfrac{13060.96}{M}](https://tex.z-dn.net/?f=%5Cdfrac%7B13060.96%7D%7BM%7D)
a= ![\dfrac{13060.96}{4410}](https://tex.z-dn.net/?f=%5Cdfrac%7B13060.96%7D%7B4410%7D)
a = 2.96 m/s²
velocity = \dfrac{2.96}{2.93}
v = 1.01 m/s
Answer:
![P=1362\ W](https://tex.z-dn.net/?f=P%3D1362%5C%20W)
is time required to heat to boiling point form initial temperature.
Explanation:
Given:
initial temperature of water, ![T_i=18^{\circ}C](https://tex.z-dn.net/?f=T_i%3D18%5E%7B%5Ccirc%7DC)
time taken to vapourize half a liter of water, ![t=18\ min=1080\ s](https://tex.z-dn.net/?f=t%3D18%5C%20min%3D1080%5C%20s)
desity of water, ![\rho=1\ kg.L^{-1}](https://tex.z-dn.net/?f=%5Crho%3D1%5C%20kg.L%5E%7B-1%7D)
So, the givne mass of water, ![m=1\ kg](https://tex.z-dn.net/?f=m%3D1%5C%20kg)
enthalpy of vaporization of water, ![h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}](https://tex.z-dn.net/?f=h_%7Bfg%7D%3D2256.4%5Ctimes%2010%5E%7B-3%7D%5C%20J.kg%5E%7B-1%7D)
specific heat of water, ![c=4180\ J.kg^{-1}.K^{-1}](https://tex.z-dn.net/?f=c%3D4180%5C%20J.kg%5E%7B-1%7D.K%5E%7B-1%7D)
Amount of heat required to raise the temperature of given water mass to 100°C:
![Q_s=m.c.\Delta T](https://tex.z-dn.net/?f=Q_s%3Dm.c.%5CDelta%20T)
![Q_s=1\times 4180\times (100-18)](https://tex.z-dn.net/?f=Q_s%3D1%5Ctimes%204180%5Ctimes%20%28100-18%29)
![Q_s=342760\ J](https://tex.z-dn.net/?f=Q_s%3D342760%5C%20J)
Now the amount of heat required to vaporize 0.5 kg of water:
![Q_v=m'\times h_{fg}](https://tex.z-dn.net/?f=Q_v%3Dm%27%5Ctimes%20h_%7Bfg%7D)
where:
mass of water vaporized due to boiling
![Q_v=0.5\times 2256.4](https://tex.z-dn.net/?f=Q_v%3D0.5%5Ctimes%202256.4)
![Q_v=1.1282\times 10^{6}\ J](https://tex.z-dn.net/?f=Q_v%3D1.1282%5Ctimes%2010%5E%7B6%7D%5C%20J)
Now the power rating of the boiler:
![P=\frac{Q_s+Q_v}{t}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BQ_s%2BQ_v%7D%7Bt%7D)
![P=\frac{342760+1128200}{1080}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B342760%2B1128200%7D%7B1080%7D)
![P=1362\ W](https://tex.z-dn.net/?f=P%3D1362%5C%20W)
Now the time required to heat to boiling point form initial temperature:
![t'=\frac{Q_s}{P}](https://tex.z-dn.net/?f=t%27%3D%5Cfrac%7BQ_s%7D%7BP%7D)
![t'=\frac{342760}{1362}](https://tex.z-dn.net/?f=t%27%3D%5Cfrac%7B342760%7D%7B1362%7D)