Answer:
23.96 N
Explanation:
From the question given above, the following data were obtained:
Mass of Chihuahua (m) = 3.63 kg
Velocity (v) = 3.3m/s
Time (t) = 0.50 s
Force (F) =?
Next, we shall determine the acceleration of the Chihuahua. This can be obtained as follow:
Velocity (v) = 3.3m/s
Time (t) = 0.50 s
Acceleration (a) =?
a = v/t
a = 3.3/0.5
a = 6.6 m/s²
Thus, the acceleration of the Chihuahua is 6.6 m/s².
Finally, we shall determine the force need to stop the Chihuahua as shown below:
Mass of Chihuahua (m) = 3.63 kg
Acceleration (a) = 6.6 m/s².
Force (F) =?
F = ma
F = 3.63 × 6.6
F = 23.96 N
Therefore, a force of 23.96 N is needed to stop the Chihuahua.
Answer:
The work done by the applied force is 259.22 J.
Explanation:
The work done by the applied force is given by:
Where:
F: is the applied horizontal force = 108.915 N
d: is the distance = 2.38 m
Hence, the work is:
Therefore, the work done by the applied force is 259.22 J.
I hope it helps you!
Answer:
huh? do you need help on math?
Explanation:
what do you mean?
Answer:
velocity = 1527.52 ft/s
Acceleration = 80.13 ft/s²
Explanation:
We are given;
Radius of rotation; r = 32,700 ft
Radial acceleration; a_r = r¨ = 85 ft/s²
Angular velocity; ω = θ˙˙ = 0.019 rad/s
Also, angle θ reaches 66°
So, velocity of the rocket for the given position will be;
v = rθ˙˙/cos θ
so, v = 32700 × 0.019/ cos 66
v = 1527.52 ft/s
Acceleration is given by the formula ;
a = a_r/sinθ
For the given position,
a_r = r¨ - r(θ˙˙)²
Thus,
a = (r¨ - r(θ˙˙)²)/sinθ
Plugging in the relevant values, we obtain;
a = (85 - 32700(0.019)²)/sin66
a = (85 - 11.8047)/0.9135
a = 80.13 ft/s²
