5.972 × 10^24 kg
it is the weight of earth
hope it is helpful to you
I have no idea what you are trying to ask sorry
<span>2002 seconds, or 33 minutes, 22 seconds.
First, let's calculate how many joules it will take to lift 78 kg against gravity for 1100 meters. So:
78 kg * 9.8 m/s^2 * 1100 m = 840840 kg*m^2/s^2
Now a watt is defined as kg*m^2/s^3, so a division of the required joules should give us a convenient value of seconds. So:
840840 kg*m^2/s^2 / 420 kg*m^2/s^3 = 2002 seconds.
And 2002 seconds is the same as 33 minutes, 22 seconds.</span>
Answer:
initial velocity = 0 m/s
final velocity = 4.92 m/s
constant acceleration so,
(a) average velocity =
(initial velocity + final velocity)/2
(b) distance = average velocity x time
substitute and calculate
Explanation:
HOOE ITS HELP ;)
Answer:
a) v = 54.7m/s
b) v = (58 - 1.66a) m/s
c) t = 69.9 s
d) v = -58.0 m/s
Explanation:
Given;
The height equation of the arrow;
H = 58t - 0.83t^2
(a) Find the velocity of the arrow after two seconds. m/s;
The velocity of the arrow v can be given as dH/dt, the change in height per unit time.
v = dH/dt = 58 - 2(0.83t) ......1
At t = 2 seconds
v = dH/dt = 58 - 2(0.83×2)
v = 54.7m/s
(b) Find the velocity of the arrow when t = a. m/s
Substituting t = a into equation 1
v = 58 - 2(0.83×a)
v = (58 - 1.66a) m/s
(c) When will the arrow hit the surface? (Round your answer to one decimal place.) t = s
the time when H = 0
Substituting H = 0, we have;
H = 58t - 0.83t^2 = 0
0.83t^2 = 58t
0.83t = 58
t = 58/0.83
t = 69.9 s
(d) With what velocity will the arrow hit the surface? m/s
from equation 1;
v = dH/dt = 58 - 2(0.83t)
Substituting t = 69.9s
v = 58 - 2(0.83×69.9)
v = -58.0 m/s
