Based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the h
igher track? _________
a. cars travel faster on the lower track
b. cars travel faster on the higher track
c. There is no difference in speed
How can the reasoning for the above answer be best explained? On the higher track, the elapsed time is__________
a. Shorter
b. Longer
c. Equal
Calculate speeds for each track. How much faster was the car on the higher track than the lower track?__________
a. .56 m/s
b. 1.54 m/s
c. 3.50 m/s
a. cars travel faster on the lower track
b. cars travel faster on the higher track
c. There is no difference in speed
How can the reasoning for the above answer be best explained? On the higher track, the elapsed time is__________
a. Shorter
b. Longer
c. Equal
Calculate speeds for each track. How much faster was the car on the higher track than the lower track?__________
a. .56 m/s
b. 1.54 m/s
c. 3.50 m/s
2 answers:
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Answer:
The energy of the system is 15 J.
Explanation:
Given that,
Energy E = 2.5 J
Amplitude = 10 cm
We need to calculate the spring constant
Using formula of mechanical energy of the system
Put the value into the formula
If the block is replaced by a block with twice the mass of the original block
Amplitude = 6 cm
We need to calculate the energy
Using formula of mechanical energy
Put the value into the formula
Hence, The energy of the system is 15 J.
Answer:
The total time for the race is 11.6 seconds
Explanation:
The parameters given are;
Total distance ran by Willie = 100.0 m
Initial acceleration = 2.00m/s²
Top speed reached with initial acceleration = 12.0 m/s
Point where Willie start to fade and decelerate = 16.0 m from the finish line
Speed with which Willie crosses the finish line = 8.00 m/s
The time and distance covered with the initial acceleration are found using the following equations of motion;
v = u₀ + a·t
v² = u₀² + 2·a·s
Where:
v = Final velocity reached with the initial acceleration = 12.0 m/s
u₀ = Initial velocity at the start of the race = 0 m/s
t = Time during acceleration
a = Initial acceleration = 2.00 m/s²
s = Distance covered during the period of initial acceleration
From, v = u₀ + a·t, we have;
12 = 0 + 2×t
t = 12/2 = 6 seconds
From, v² = u₀² + 2·a·s, we have;
12² = 0² + 2×2×s
144 = 4×s
s = 144/4 =36 meters
Given that the Willie maintained the top speed of 12.0 m/s until he was 16.0 m from the finish line, we have;
Distance covered at top speed = 100 - 36 - 16 = 48 meters
Time, of running at top speed = Distance/velocity = 48/12 = 4 seconds
The deceleration from top speed to crossing the line is found as follows;
v₁² = u₁² + 2·a₁·s₁
Where:
u₁ = v = 12 m/s
v₁ = The speed with which Willie crosses the line = 8.00 m/s
s₁ = Distance covered during decelerating = 16.0 m
a₁ = Deceleration
From which we have;
8² = 12² + 2 × a × 16
64 = 144 + 32·a
64 - 144 = 32·a
32·a = -80
a = -80/32 = -2.5 m/s²
From, v₁ = u₁ + a₁·t₁
Where:
t₁ = Time of deceleration
We have;
8 = 12 + (-2.5)·t₁
t₁ = (8 - 12)/(-2.5) = 1.6 seconds
The total time = t + + t₁ =6 + 4 + 1.6 = 11.6 seconds.