The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
Explanation:
Consider the vertical motion of ball,
We have equation of motion v = u + at
Initial velocity, u = u sin θ
Final velocity, v = 0 m/s
Acceleration = -g
Substituting
v = u + at
0 = u sin θ - g t
This is the time of flight.
Consider the horizontal motion of ball,
Initial velocity, u = u cos θ
Acceleration, a =0 m/s²
Time,
Substituting
s = ut + 0.5 at²
This is the range.
In this problem
u = 30 m/s
g = 9.81 m/s²
θ = 45° - For maximum range
Substituting
Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.
So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground
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
Answer:
Vf= 7.29 m/s
Explanation:
Two force act on the object:
1) Gravity
2) Air resistance
Upward motion:
Initial velocity = Vi= 10 m/s
Final velocity = Vf= 0 m/s
Gravity acting downward = g = -9.8 m/s²
Air resistance acting downward = a₁ = - 3 m/s²
Net acceleration = a = -(g + a₁ ) = - ( 9.8 + 3 ) = - 12.8 m/s²
( Acceleration is consider negative if it is in opposite direction of velocity )
Now
2as = Vf² - Vi²
⇒ 2 * (-12.8) *s = 0 - 10²
⇒-25.6 *s = -100
⇒ s = 100/ 25.6
⇒ s = 3.9 m
Downward motion:
Vi= 0 m/s
s = 3.9 m
Gravity acting downward = g = 9.8 m/s²
Air resistance acting upward = a₁ = - 3 m/s²
Net acceleration = a = g - a₁ = 9.8 - 3 = 6.8 m/s²
Now
2as = Vf² - Vi²
⇒ 2 * 6.8 * 3.9 = Vf² - 0
⇒ Vf² = 53. 125
⇒ Vf= 7.29 m/s
Explanation:
Given that,
Initial speed of the bag, u = 7.3 m/s
Height above ground, s = 24 m
We need to find the speed of the bag just before it reaches the ground. It can be calculated using third equation of motion as :
v = 22.88 m/s
So, the speed of the bag just before it reaches the ground is 22.38 m/s. Hence, this is the required solution.
Answer:
it Give only one of them a positive or negative charge
