Answer:
a = 120 m/s²
Explanation:
We apply Newton's second law in the x direction:
∑Fₓ = m*a Formula (1)
Known data
Where:
∑Fₓ: Algebraic sum of forces in the x direction
F: Force in Newtons (N)
m: mass (kg)
a: acceleration of the block (m/s²)
F = 1200N
m = 10 kg
Problem development
We replace the known data in formula (1)
1200 = 10*a
a = 1200/10
a = 120 m/s²
Answer:
a)54L/min
b)0.845
Explanation:
a) A x V=![A_1V_1+ A_2V_2+A_3V_3](https://tex.z-dn.net/?f=A_1V_1%2B%20A_2V_2%2BA_3V_3)
where suffix 1,2,3 refers to the three pipes.
=27L/min+16L/min+11 L/min
=54L/min
b) A x V=54L/min =>
x v
d= 2 cm
x v = 54
v=
x
->
x
=27L/min =>
x ![v_1](https://tex.z-dn.net/?f=v_1)
= 1.3cm
x
= 27
=
x ![\frac{27}{1.3^2}](https://tex.z-dn.net/?f=%5Cfrac%7B27%7D%7B1.3%5E2%7D)
Next is to find the ratio of speed i.e ![\frac{v}{v_1}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%7D%7Bv_1%7D)
x
/
x
=>
![\frac{1.3^2}{2^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1.3%5E2%7D%7B2%5E2%7D)
= 0.845
Answer:
14 rev
Explanation:
= initial angular velocity = 2.5 revs⁻¹
= final angular velocity = 0.8 revs⁻¹
= Angular acceleration = - 0.2 revs⁻²
= Angular displacement
Using the equation
![w^{2} = w_{o}^{2} + 2 \alpha \theta\\0.8^{2} = 2.5^{2} + 2 (- 0.2) \theta\\ \theta = 14 rev](https://tex.z-dn.net/?f=w%5E%7B2%7D%20%3D%20w_%7Bo%7D%5E%7B2%7D%20%2B%202%20%5Calpha%20%5Ctheta%5C%5C0.8%5E%7B2%7D%20%3D%202.5%5E%7B2%7D%20%2B%202%20%28-%200.2%29%20%5Ctheta%5C%5C%20%5Ctheta%20%3D%2014%20rev)
So the number of revolutions are 14
KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J
Answer:
False
Explanation:
Because when you go through east
( +x axis ) then you go to west ( -x axis )
You will subtract -9 from +15
it's become +6
( I talk about the displacement not distance) ( West = - East )
I hope that it's a clear ") .