The answer is c hope it helps
Answer:
1.92 J
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 200 Kg
Spring constant (K) = 10⁶ N/m
Workdone =?
Next, we shall determine the force exerted on the spring. This can be obtained as follow:
Mass (m) = 200 Kg
Acceleration due to gravity (g) = 9.8 m/s²
Force (F) =?
F = m × g
F = 200 × 9.8
F = 1960 N
Next we shall determine the extent to which the spring stretches. This can be obtained as follow:
Spring constant (K) = 10⁶ N/m
Force (F) = 1960 N
Extention (e) =?
F = Ke
1960 = 10⁶ × e
Divide both side by 10⁶
e = 1960 / 10⁶
e = 0.00196 m
Finally, we shall determine energy (Workdone) on the spring as follow:
Spring constant (K) = 10⁶ N/m
Extention (e) = 0.00196 m
Energy (E) =?
E = ½Ke²
E = ½ × 10⁶ × (0.00196)²
E = 1.92 J
Therefore, the Workdone on the spring is 1.92 J
Answer: The density of this piece of jewelry is 
Explanation:
To calculate the density, we use the equation:
Mass of piece of jewellery = 130.8 g
Density of piece of jewellery = ?
Volume of piece of jewellery =( 62.4-47.7 ) ml = 14.7 ml = 
Putting values in above equation, we get:
Thus density of this piece of jewelry is
Answer: 0.25 m/s
Explanation: Speed = wavelengt · frequency
v = λf and frequency is 1/period f = 1/T
Then v = λ/T = 5 m / 20 s = 0.25 m/s
Answer:
160N
Explanation: When 80kg mass is one group . It's reaction force acting on a ground.
Weight of the object = 80*10
= 800 N
Here we are given cofficient of static friction its 0.2. It should be smaller than 1
Friction force = Reaction * Friction Cofficient
Reaction = 800N ( Considering Vertical Equilibrium )
F = 800* 0.2
F = 160N
