negative acceleration- deceleration
Answer:
a

b
The value is 
Explanation:
From the question we are told that
The mass is
The spring constant is 
The instantaneous speed is 
The position consider is x = 0.750A meters from equilibrium point
Generally from the law of energy conservation we have that
The kinetic energy induced by the hammer = The energy stored in the spring
So

Here a is the amplitude of the subsequent oscillations
=> 
=> 
=> 
Generally from the law of energy conservation we have that
The kinetic energy by the hammer = The energy stored in the spring at the point considered + The kinetic energy at the considered point

=> 
=> 

- Speed of the mobile = 250 m/s
- It starts decelerating at a rate of 3 m/s²
- Time travelled = 45s

- Velocity of mobile after 45 seconds

We can solve the above question using the three equations of motion which are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

We are provided with,
- u = 250 m/s
- a = -3 m/s²
- t = 45 s
By using 1st equation of motion,
⇛ v = u + at
⇛ v = 250 + (-3)45
⇛ v = 250 - 135 m/s
⇛ v = 115 m/s
✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Explanation:
Given:
v₀ = 0 m/s
a = 3 m/s²
t = 4 s
Find: Δx and v
Δx = v₀ t + ½ at²
Δx = (0 m/s) (4 s) + ½ (3 m/s²) (4 s)²
Δx = 24 m
v = at + v₀
v = (3 m/s²) (4 s) + 0 m/s
v = 12 m/s
Given:
Sample 1:
Chloroform is 
12 g Carbon
1.01 g Hydrogen
106.4 g Cl
Sample 2:
30.0 g of Carbon
Solution:
mass of chloroform from sample 1:
12 + 1.01 +106.4 =119.41 g
Now, for the total mass of chloroform in sample 2:
mass of chloroform 

mass of chloroform = 119.41 
= 298.53 g