The protons have a positive charge.
When a plane travels 395,000 meters in 9,000 seconds its speed is 395,000 ÷ 9000 = 43.888888 (the 8 repeats) meters per second.
Answer:
The average velocity is
and respectively.
Explanation:
Let's start writing the vertical position equation :
Where distance is measured in meters and time in seconds.
The average velocity is equal to the position variation divided by the time variation.
= Δx / Δt =
For the first time interval :
t1 = 5 s → t2 = 8 s
The time variation is :
For the position variation we use the vertical position equation :
Δx = x2 - x1 = 1049 m - 251 m = 798 m
The average velocity for this interval is
For the second time interval :
t1 = 4 s → t2 = 9 s
Δx = x2 - x1 = 1495 m - 125 m = 1370 m
And the time variation is t2 - t1 = 9 s - 4 s = 5 s
The average velocity for this interval is :
Finally for the third time interval :
t1 = 1 s → t2 = 7 s
The time variation is t2 - t1 = 7 s - 1 s = 6 s
Then
The position variation is x2 - x1 = 701 m - (-1 m) = 702 m
The average velocity is
Answer:
the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg
Explanation:
Given the data in the question;
first we compute the total volume of mud
V = ( 1.9×1000)m × (0.55×1000)m × (2.4×1000)m
V = 2.508 × 10⁹ m³
same volume of mud spread uniformly over 1.4 × 1.4 km² area
Hence,
V = (1.4×1000)m × (1.4×1000)m × depth
2.508 × 10⁹ = 1400m × 1400m × depth
2.508 × 10⁹ m³ = 1960000m² × depth
depth = 2.508 × 10⁹ m³ / 1960000 m²
depth = 1279.59 m
so volume of the mud sitting above 3.5 m² area of the valley floor will be;
⇒ 3.5 m² × 1279.59 m
⇒ 4478.565 m³
so mass will be
m = 4478.565 × 1900 kg
m = 8509273.5 kg
Therefore, the mass of the mud sitting above a 3.5 m2 area of the valley floor is 8509273.5 kg