Answer:
v₂ = 22.5 m/s
Explanation:
Given that
For puck 1
m₁= 1 kg
u₁= 30 m/s
For puck 2
m₂= 1 kg
u₂= 0 m/s
After collision
Puck 1 have velocity v₁=7.5 m/s
Take puck 2 will have velocity v₂
From linear momentum conservation
P₁=P₂
m₁ u₁+m₂ u₂=m₁ v₁+m₂ v₂
1 x 30 + 1 x 0 = 1 x 7.5 + 1 x v₂
30 - 7.5 =v₂
v₂ = 22.5 m/s
Answer:
v_f = 6.92 x 10^(4) m/s
Explanation:
From conservation of energy,
E = (1/2)mv² - GmM/r
Where M is mass of sun
Thus,
E_i = E_f will give;
(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)
m will cancel out to give ;
(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)
Let's make v_f the subject;
v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]
G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²
Mass of sun is 1.9891 x 10^(30) kg
v_i = 2.1×10⁴ m/s
r_i = 2.5 × 10^(11) m
r_f = 4.9 × 10^(10) m
Plugging in all these values, we have;
v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12
v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]
v_f = √[(441000000) + (435.38 x 10^(7))
v_f = 6.92 x 10^(4) m/s
Answer:
a diagonal line on the graph
Explanation:
the slope would be constant at an increasing rate
y=x
uniform= increasing at a constant rate
Answer:
False. the system does not complete the circle movement
Explanation:
For this exercise, we must find the rope tension at the highest point of the path,
-T - W = m a
The acceleration is centripetal
a = v² / R
T = ma - mg
T = m (v² / R - g)
The minimum tension that the rope can have is zero (T = 0)
v² / R - g = 0
v = √ g R
Let's find out what this minimum speed is
v = √ 9.8 1
v = 3.13 m / s
We see that the speed of the body is less than this, so the system does not complete the movement.
