Deceleration—the ability to slow down and control force production—is often ignored during training; but deceleration technique is critical for most sports. Speed is often the factor that separates the elite from the average athlete. Credit source is stack.com have a nice day! I told you it was from the internet in case you couldn’t use the internet Hope this helped :)!
Answer:
Shown by explanation;
Explanation:
The heat of the sample = mass ×specific heat capacity of the sample × temperature change(∆T)
Assumption;I assume the mass of the samples are : 109g and 192g
∆T= 30.1-21=8.9°c.
The heat of the samples are for 109g are:
0.109 × 4186 × 8.9 =4060.84J
For 0.192g are;
∆T= 67-30.1-=36.9°c
0.192 × 4186×36.9=29656.97J
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
I think it false. Sorry if i'm wrong.