Answer:
- The procedure is: solve the quadratic equation for
.
Explanation:
This question assumes uniformly accelerated motion, for which the distance d a particle travels in time t is given by the general equation:
That is a quadratic equation, where the independent variable is the time .
Thus, the procedure that will find the time t at which the distance value is known to be D is to solve the quadratic equation for .
To solve it you start by changing the equation to the general form of the quadratic equations, rearranging the terms:
Some times that equation may be solved by factoring, and always it can be solved by using the quadratic formula:
Where:
That may have two solutions. Some times one of the solution makes no physical sense (for example time cannot be negative) but others the two solutions are valid.
Answer:
The second ball lands 1.5 s after the first ball.
Explanation:
Given;
initial velocity of the ball, u = 12 m/s
height of fall, h = 35 m
initial velocity of the second, v = 12 m/s
Time taken for the first ball to land;
determine the maximum height reached by the second ball;
v² = u² -2gh
at maximum height, the final velocity, v = 0
0 = 12² - (2 x 9.8)h
19.6h = 144
h = 144 / 19.6
h = 7.35 m
time to reach this height;
Total height above the ground to be traveled by the second ball is given as;
= 7.35 m + 35m
= 42.35 m
Time taken for the second ball to fall from this height;
total time spent in air by the second ball;
T = t₁ + t₂
T = 1.23 s + 2.94 s
T = 4.17 s
Time taken for the second ball to land after the first ball is given by;
t = 4.17 s - 2.67 s
T = 1.5 s
Therefore, the second ball lands 1.5 s after the first ball.
Answer:
e% = 3.4%
Explanation:
This is a calorimetry problem where the heat released equals the heat absorbed
m 
(T₀ - T_f) = M c_{e2} (T₁ - T_f)
Index 1 refers to water and index 2 to metal, in this case it asks for the specific heat of the metal (c_{e2})
c_{e2} = m / M c_{e1} (T_f -T₀) / (T₁ - T_f)
Let's calculate
c_{e} = 60/100 4.19 (24-20) / (100-24)
c_{e2} = 0.1323 j / gC
This metal is possibly lead, which is its specific heat is 0.128 J / gC
The percentage error is
e% = (c_{e2} - 0.128) /0.128 100
e% = 3.4%
