Answer:
108 extended days
Explanation:
Regular school hours a day = 6 hr
No. of school days to make up by extending the regular hours = 3 days
Amount of time added to the regular hours of school = 10 min
No. of extended school days to make up the 3 school days by following the above mentioned criteria be x.
Time of school hours in 3 days = 
are required to make up 3 days of school having 6 hours of regular timing with 10 minutes of add-on time each day.
Answer:
3.07 m/s
Explanation:
We know that from kinematics equation
and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity
Making u the subject then
Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then
To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.
PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation
Where,
G = Gravitational Universal Constant
d = Distance
M = Mass
Radius earth center of mass
PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,
PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is
At the same time we have that centripetal acceleration is given as
Replacing
Answer:
C
Explanation:
Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces
I hope this helps a little bit
Answer:
Because of the speed of the sound.
Explanation:
The first thing that happens in such cases is to take into account the speed of the sound. First, we see that the player hits the ball with the bat, if we are in the stands far enough we will hear the sound of the batting time later, this time depends on the speed of the sound which is equal to 345 [m/s].
Another visible and practical example is a fireworks display, where people nearby immediately hear the explosion. while those at a great distance will be able to see first the explosion followed by the sound.
With the following equation, we can calculate how long it takes to hear a hit or explosion
t = x / v
where:
x = distance [m]
v = sound velocity = 345 [m/s]
t = time [s]
