1. Frequency 2. measure from trough to trough
Answer:
(2)
Explanation:
<u>a)Kinematics equation for the first ball:</u>
initial position is the building height
The ball reaches the ground, y=0, at t=t1:
(1)
Kinematics equation for the second ball:
initial position is the building height
the ball is dropped
The ball reaches the ground, y=0, at t=t2:
(2)
the second ball is dropped a time of 1.03s later than the first ball:
t2=t1-1.03 (3)
We solve the equations (1) (2) (3):
vo=8.9m/s
t2=t1-1.03 (3)
t2=3.29sg
(2)
b)
t1 must : t1>1.03 and t1>0
limit case: t1>1.03:
limit case: t1>0:
A calculator must be used. To put your calculator in degree mode, press the MODE button and select degree, the press the 2nd button then MODE again. For most TI calculators, press the 2nd button then press the cos button then enter the value 0.34. This will give you an answer of 70.123 (when you round to 3 decimal places).
The answer is C
Answer:
a. v₁ = 16.2 m/s
b. μ = 0.251
Explanation:
Given:
θ = 15 ° , r = 100 m , v₂ = 15.0 km / h
a.
To determine v₁ to take a 100 m radius curve banked at 15 °
tan θ = v₁² / r * g
v₁ = √ r * g * tan θ
v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s
b.
To determine μ friction needed for a frightened
v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg
v₂ = 4.2 m/s
fk = μ * m * g
a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²
a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²
F₁ = m * a₁ , F₂ = m * a₂
fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)
μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g
μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)
μ = 0.251
Answer:
So we have a measure in grams, we can start with something like:
143.523 grams.
Now we want this measurement to be precise to the nearest tenth of a gram.
The nearest tenth of a gram is the first digit after the decimal point, then the digits that come after this are not useful, because they are outside our precision range.
Then we must write our measurement as:
143.5 grams
Where the digit that came after the 5 was a 2, so we rounded down.
