Answer:
The football leaves with the velocity, u = 15.68 m/s
Explanation:
Given data,
The football bounces back up off the ground and is airborne for, t = 3.2 s
Let the football bounces back up off the ground in the vertical direction
The formula for time of flight is given by,
t = 2u /g
∴ u = gt / 2
Substituting the values,
u = 9.8 x 3.2 / 2
u = 15.68 m/s
Hence, the football leaves with the velocity, u = 15.68 m/s
Answer: magnitude and direction
Explanation:They are the two aspects of force that scientists measure
4 times as much distance to stop
Answer:
r = 0.5 m
Explanation:
First we find the angular speed of the ball by using its period:
ω = θ/t
For the time period:
ω = angular speed = ?
θ = angular displacement = 2π rad
t = time period = 0.5 s
Therefore,
ω = 2π rad/0.5 s
ω = 12.56 rad/s
Now, for the radius:
v = rω
r = v/ω
where,
v = linear speed = 6.29 m/s
r = radius = ?
r = (6.29 m/s)/(12.56 rad/s)
<u>r = 0.5 m</u>
Answer:
a) a geostationary satellite is that it is always at the same point with respect to the planet,
b) f = 2.7777 10⁻⁵ Hz
c) d) w = 1.745 10⁻⁴ rad / s
Explanation:
a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet
- T = 10 h (3600 s / 1h) = 3.6 104 s
b) the period the frequency are related
T = 1 / f
f = 1 / T
f = 1 / 3.6 104
f = 2.7777 10⁻⁵ Hz
c) the distance traveled by the satellite in 1 day
The distance traveled is equal to the length of the circumference
d = 2pi (R + r)
d = 2pi (69 911 103 + 120 106)
d = 1193.24 m
d) the angular velocity is the angle traveled between the time used.
.w = 2pi /t
w = 2pi / 3.6 10⁴
w = 1.745 10⁻⁴ rad / s
how fast is
v = w r
v = 1.75 10-4 (69.911 106 + 120 106)
v = 190017 m / s
