After x seconds, an object will fall
where a is acceleration due to gravity and t is time
so when t=3.3
the distance it will fall is
=53.361m
it will fal 53.361 meters
Answer:
Explanation:
First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:
The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.
Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:
The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:
Answer:
32 amu is the right choice because both protons and neutrons have a mass of 1 amu. Electrons have no mass so go with the last choice
Answer:
The answer should be light or other electromagnetic radiation
Explanation:
Such as x-rays or other things like that.
<em>-Hope This Helps!</em>
<em>-Justin:)</em>
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>