The velocity of the package after it has fallen for 3.0 s is 29.4 m/s
From the question,
A small package is dropped from the Golden Gate Bridge.
This means the initial velocity of the package is 0 m/s.
We are to calculate the velocity of the package after it has fallen for 3.0 s.
From one of the equations of kinematics for objects falling freely,
We have that,
v = u + gt
Where
v is the final velocity
u is the initial velocity
g is the acceleration due to gravity
and t is time
To calculate the velocity of the package after it has fallen for 3.0 s
That means, we will determine the value of v, at time t = 3.0 s
The parameters are
u = 0 m/s
g = 9.8 m/s²
t = 3.0 s
Putting these values into the equation
v = u + gt
We get
v = 0 + (9.8×3.0)
v = 0 + 29.4
v = 29.4 m/s
Hence, the velocity of the package after it has fallen for 3.0 s is 29.4 m/s
Learn more here: brainly.com/question/13327816
Answer:
Before fueling the tank.
Explanation:
All the stated concerns such as the condition of fuel line and fuel vents should always be checked before fueling the tank.
While fueling the tank make sure not to fill it up to brim. Doing this may gets to blast as fuel when heated will not have any space to expand. Fuel vent must be in good condition as it release vacuum and pressure build inside the tank.
Fuel lines helps to move to move fuel from one place to another, hence any damage or leakage in this can result into a big collapse so all these must be properly checked.
Answer:
The answer is a. littoral
A high back is where the top proportion of the seat is all in one piece.(no adjustable headrests)Generally a low back is where the seat has an adjustable headrest
Answer:
a. 127 V b 311 V
Explanation:
a. The RMS value of E(t) = 1/T∫[E(t)]². E(t) = 180 sin(200πt). Since the frequency f = 100 cycles per second, the period, T = 1/f = 1/100 = 0.01 s.
So, 1/T∫[E(t)]² = 1/T∫[180 sin(200πt)]² = 180²/0.01∫sin(200πt)]²
Using trigonometric identity sin²Ф = (1 - cos2Ф)/2 where Ф = ωt
1/T∫[E(t)]² = 180²/0.01∫(1 - cos2Ф)/2. We integrate from 0 to T, , we have
1/T∫[E(t)]² = 180²/(0.01 × 2)(t - sin2ωt/2ω)
1/T∫[E(t)]² = 180²/(0.02[(T - (sin2π)/(2 × 200π) ) - (0 - [sin(2 × 0)]/(2 × 200π))
1/T∫[E(t)]² = 180²/(0.02)[(0.01 - 0)
1/T∫[E(t)]² = 180²/2
E(t)RMS = √1/T∫[E(t)]²
= √180²/2
= 180/√2
= 127.28 V
= 127 V to the nearest whole number
b. Since E(t)(RMS) = A/√2 where A = voltage amplitude and E(t)(RMS) = 220 V,
A = √2E(t)(RMS) =
√2 × 220 V
= 311.13 V
= 311 V to the nearest whole number
