Answer:
Volume of gasoline that expands and spills out is 1.33 ltr
Explanation:
As we know that when temperature of the liquid is increased then its volume will expand and it is given as

here we know that

volume expansion coefficient of the gasoline is given as

change in temperature is given as


Now we have


Answer:
50%
Explanation:
Humidity is the amount water vapor present in the atmosphere.
Relative humidity is defined as the ratio of partial water vapor present in air to the actual water vapor at a particular temperature. It is expressed in percentage and the higher the percentage RH, the more the saturated water vapor present in the atmosphere and vice versa.
It is expressed mathematically as shown;
RH = actual water vapor in air/saturated water vapor × 100%
If the actual water vapor in the air was 4 grams per cubic meter and the air's capacity to hold water vapor was 8 grams per cubic meter
Actual water vapor = 4g/cm³
Air's water capacity (saturated water vapor) = 8g/cm³
RH = 4/8×100
RH = 50%
Answer:
Explanation:
(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.
(b) Yes, it does and the formula for force of gravity between any two object is
F = G
where m1 and m2 are masses of the first and second object respectively
r is the distance between the center of the two masses
G is the gravitational constant
Answer:
it can be calculated by measuring the final distance away from a point, and then subtracting the initial distance
Answer:
a) v = 19,149.6 m/s
b) f = 95%
c) t = 346.5min
Explanation:
First put all values in metric units:

The equation of motion you need is:
where
is the final velocity, a is acceleration and t is time in hours.
Since the spaceship starts from 0 velocity:

Next, you need to calculate the distances traveled on each interval, considering that both starting and final intervals travel the same distance because the acceleration and time are equal. For this part you need the next motion equation:

solving for first and last interval:
Since the spaceship starts and finish with 0 velocity:

Then the ship traveled
at constant speed, which means that it traveled:

Which in percentage is 95% of the trip.
to calculate total time you need to calculate the time used during constant speed:

That added to the other interval times:
