Probably not what you were expecting... the average bottle of water is 24 ounces. 5 milliliters is about the amount of water in a spoon. Hope this helps!!!
Answer:
20.96 m/s^2 (or 21)
Explanation:
Using the formula (final velocity - initial velocity)/time = acceleration, we can plug in values and manipulate the problem to give us the answer.
At first, we know a car is going 8 m/s, that is its initial velocity.
Then, we know the acceleration, which is 1.8 m/s/s
We also know the time, 7.2 second.
Plugging all of these values in shows us that we need to solve for final velocity. We can do so by manipulating the formula.
(final velocity - initial velocity) = time * acceleration
final velocity = time*acceleration + initial velocity
After plugging the found values in, we get 20.96 m/s/s, or 21 m/s
Even with no friction, it depends on the slope of the roof. That is, it depends on how much elevation (altitude) he loses during the slide.
Whatever that number is ... call it 'h' ... Santa's speed when he reaches the edge is
Square root of (19.6h) meters per second.
It doesn't matter how much he weighs, or how far he has slud. Only how much altitude he lost on the slope while sliding.
Answer:
2.23 × 10^6 g of F- must be added to the cylindrical reservoir in order to obtain a drinking water with a concentration of 0.8ppm of F-
Explanation:
Here are the steps of how to arrive at the answer:
The volume of a cylinder = ((pi)D²/4) × H
Where D = diameter of the cylindrical reservoir = 2.02 × 10^2m
H = Height of the reservoir = 87.32m
Therefore volume of cylindrical reservoir = (3.142×202²/4)m² × 87.32m = 2798740.647m³
1ppm = 1g/m³
0.8ppm = 0.8 × 1g/m³
= 0.8g/m³
Therefore to obtain drinking water of concentration 0.8g/m³ in a reservoir of volume 2798740.647m³, F- of mass = 0.8g/m³ × 2798740.647m³ = 2.23 × 10^6 g must be added to the tank.
Thank you for reading.
Answer:
25.6 m/s
Explanation:
Draw a free body diagram of the sled. There are two forces acting on the sled:
Normal force pushing perpendicular to the hill
Weight force pulling straight down
Take sum of the forces parallel to the hill:
∑F = ma
mg sin θ = ma
a = g sin θ
a = (9.8 m/s²) (sin 38.0°)
a = 6.03 m/s²
Given:
v₀ = 0 m/s
a = 6.03 m/s²
t = 4.24 s
Find: v
v = at + v₀
v = (6.03 m/s²) (4.24 s) + (0 m/s)
v = 25.6 m/s
