Answer:
The angle is 
Explanation:
From the question we are told that
The distance of the dartboard from the dart is 
The time taken is 
The horizontal component of the speed of the dart is mathematically represented as

where u is the the velocity at dart is lunched
so

substituting values

=> 
From projectile kinematics the time taken by the dart can be mathematically represented as

=> 


=> 
![\theta = tan^{-1} [0.277]](https://tex.z-dn.net/?f=%5Ctheta%20%20%3D%20%20tan%5E%7B-1%7D%20%5B0.277%5D)

Answer:
0.74 N/cm
Explanation:
The following data were obtained from the question:
Mass (m) = 3 Kg
Extention (e) = 40 cm
Spring constant (K) =?
Next, we shall determine the force exerted on the spring.
This can be obtained as follow:
Mass (m) = 3 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = mg
F = 3 × 9.8
F = 29.4 N
Finally, we shall determine the spring constant of the spring. This can be obtained as follow:
Extention (e) = 40 cm
Force (F) = 29.4 N
Spring constant (K) =?
F = Ke
29.4 = K × 40
Divide both side by 40
K = 29.4 / 40
K = 0.74 N/cm
Therefore, the spring constant of the spring is 0.74 N/cm
Answer:
The weight limit of 300kg is the maximum amount the machine can handle so it can be dangerous to exceed the maximum load.
<span>net work = change in kinetic energy
for Block B, we just have the force from block A acting on it
F(ab)d= .5(1)vf² - .5(1)(2²)
F(ab)d= .5vf² - 2
Block A, we have the force from the hand going in one direction and the force of block B on A going the opposite direction
10-F(ba)d = .5(4)vf² - .5(4)(2²)
10-F(ba)d = 2vf² - 8
F(ba)d = 18 - 2vf²
now we have two equations:
F(ba)d = 18 - 2vf²
F(ab)d= .5vf² - 2
since the magnitude of F(ba) and F(ab) is the same, substitute and find vf (I already took into account the direction when solving for F(ab)
10-.5vf² + 2 = 2vf² - 8
12 - .5vf² = 2vf² - 8
20 = 2.5vf²
vf² = 8
they both will have the same velocity
KE of block A= .5(4)(2.828²) = 16 J
KE of block B=.5(1)(2.828²) = 4 J</span>