A toy rocket is shot straight up into the air with an initial speed of 45.0 m/s how long does it takes for the rocket to reach i
2 answers:
Explanation:
Given that,
Initial speed of the toy rocket, u = 45 m/s
Let t is the time taken by the rocket to reach it's highest point. At the highest point the final speed of the rocket is 0. Using equation of kinematics to find the value of t as :
Here, a = -g
t = 4.59 seconds
So, the rocket will take 4.59 seconds to reach its highest point. Hence, this is the required solution.
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Answer:
The tension in the left side string = 17.21 N
The tension in the right side string = F = 27.3 N
Explanation:
Given that
F= 27.3 N
M= 1.43 kg ,r= 0.0792 m
Moment of inertia of disk ,I = 0.5 m r²
I = 0.5 x 1.43 x 0.0792² = 0.0044 kg.m²
m= 0.7 kg
Lets take linear acceleration of system is a m/s²
Lets take tension in left side string = T
From Newtons law
T- mg = ma
T- 0.7 x 10 = 0.7 a ----------1
(F - T) r = I α
α = Angular acceleration of disk
a = α r
(F - T) r = I α
(F - T) r² = I a
( 27.3 - T) x 0.0792² = 0.0044 a --------2
Form equation 1 and 2
a= 1.42 T - 10 m/s²
a = 1.42 ( 27.3 - T) m/s²
1.42 T - 10 = 38.9 - 1.42 T
T=17.21 N
The tension in the right side string = F = 27.3 N
Answer:
v = 9.04 m / s
Explanation:
For this exercise we can use the relation that the work of the non-conservative force (friction) is equal to the variation of the mechanical energy of the system.
W = Em_f - Em₀ (1)
Starting point. Lower slope
Em₀ = K = ½ m v²
highest point. Where is the skier at a height h
Em_f = U = m g h
The work of rubbing
W = -fr x
the negative sign is because the friction force opposes the movement.
Let's set a reference system where the x axis is parallel to the slope and the y axis is perpendicular
let's use trigonometry to break down the weight
cos θ = W_y / W
sin θ = Wₓ / W
W_y = W cos θ
Wₓ = W sin θ
Y axis
N - Wₓ = 0
N = mg sin θ
X axis
fr = m a
the friction force has the expression
fr = μ N
fr = μ mg sin θ
we look for the job
W = - μ mg sin θ x
where x is the distance along the slope
we substitute in 1
-μ mg sin θ x = mg h - ½ m v²
let's use trigonometry to find the distance x
tan 30 = h / x
x = h / tan 30
we substitute
- = m gh - ½ m v²
we use
tan 30 = sin30 / cos30
v² = 2g h + 2 μ g h cos 30
v =
let's calculate
v =
v = 9.04 m / s
The total work done is
<span>W=60 plus 120
=180J
For the graphs please find the attached image</span>
<span>W=60 plus 120
=180J
For the graphs please find the attached image</span>