Momentum = mass x velocity
12 = 4 x v | ÷ both sides by 4
12 ÷ 4 =v
v= 3 m/s
Answer:
p2 = 9.8×10^4 Pa
Explanation:
Total pressure is constant and PT = P = 1/2×ρ×v^2
So p1 + 1/2×ρ×(v1)^2 = p2 + 1/2×ρ×(v2)^2
from continuity we have ρ×A1×v1 = ρ×A2×v2
v2 = v1×A1/A2
and
r2 = 2×r1
then:
A2 = 4×A1
so,
v2 = (v1)/4
then:
p2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v2)^2 = p1 + 1/2×ρ×(v1)^2 - 1/2×ρ×(v1/4)^2
p2 = 3.0×10^4 Pa + 1/2×(1000 kg/m^3)×(12m/s)^2 - 1/2×(1000kg/m^3)×(12^2/16)
= 9.75×10^4 Pa
= 9.8×10^4 Pa
Therefore, the pressure in the wider section is 9.8×10^4 Pa
The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude <em>w</em>), the normal force (mag. <em>n</em>), and the friction force (maximum mag. <em>f</em> ).
In the horizontal direction, we have
<em>n</em> cos(120º) + <em>f</em> cos(30º) = 0
-1/2 <em>n</em> + √3/2 <em>f</em> = 0
<em>n</em> = √3 <em>f</em>
and in the vertical,
<em>n</em> sin(120º) + <em>f</em> sin(30º) + (-<em>w</em>) = 0
<em>n</em> sin(120º) + <em>f</em> sin(30º) = (50 kg) (9.80 m/s²)
√3/2 <em>n</em> + 1/2 <em>f</em> = 490 N
Substitute <em>n</em> = √3 <em>f</em> and solve for <em>f</em> :
√3/2 (√3 <em>f </em>) + 1/2 <em>f</em> = 490 N
2 <em>f</em> = 490 N
<em>f</em> = 245 N
(pointed up the incline)
Gravitational potential energy=mass of object x gravitational field strength on earth(9.8 usually rounded to 10) x the height the object is held at
Therefore if two objects were held at the same height, the object with more mass(the heavier object) will fall faster because it's gravitational potential energy is greater than that of the lighter object
