The correct answer is D: Watt. This unit was named after James Watt, and
is used to express the equivalent of one joule per second in energy. In
experiments and on the packaging for electrical products such as light-bulbs, the measurement will usually be written in its abbreviated
format: W.
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Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
The force is still 1 pound.
The TORQUE around the pivot is (force) x (distance from the pivot) = 1 foot-pound. (B)
Answer:0.61 m
Explanation:
Given
Initial velocity (u)=7.40 m/s
launch angle
height of fence=0.7 m
Trajectory of Projectile is given by
For x=2.1 m
At x=2.1 m Y=1.31
therefore Fox clear fence by a margin =1.31-0.7=0.61 m
True if you have proper stance and use your body the right way then the ball will be below your waist to allow for more control.