For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.
<h3>Explanation</h3>
How long does it take for the ball to reach the goal?
Let the distance between the kicker and the goal be
meters.
Horizontal velocity of the ball will always be
until it lands if there's no air resistance.
The ball will arrive at the goal in
seconds after it leaves the kicker.
What will be the height of the ball when it reaches the goal?
Consider the equation
.
For this soccer ball:
,
,
since the player kicks the ball "from ground level."
when the ball reaches the goal.
.
Solve this quadratic equation for
,
.
meters when
meters.
or
meters when
meters.
In other words,
- For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
- For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.
You have to reduce 2.00 an5.00 I order to use the×that=0.800
Answer:
Density (φ) = 0,8827 Kg/L
Specific weight (Ws) = 8,65 N/L
Specific gravity (Gs) = 0,8827 (without unit)
Explanation:
The density formula: φ =
I know the mass "m", I need to find out the volume of the cylinder (V)
V = π* r²*h
The radius "r" is equal to half the diameter (150mm) = 75mm
Now I can find out the density (φ)
φ =
= 0,8827 Kg/L
The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:
Ws = φ*g
(g = gravity = 9,8 m/s²)
Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :
Gs = φ(o) / φ(w)
(φ(w) = 1 Kg/L
Hope this can help you !!
Answer:
0 J
Explanation:
given,
mass of the ball = 5 kg
radius of the horizontal circle = 0.5 m
tension in the string = 10 N
Work done = ?
Work done by the tension in the circular path will be equal to zero.
This is because body moves in circular path, the centripetal force act along the radius of the circle and motion is right angle to the tension on the string.
so, work done = F s cos θ
θ = 90°,
work done = F s cos 90° ∵ cos 90° = 0
Work done = 0 J
Answer:
970 kN
Explanation:
The length of the block = 70 mm
The cross section of the block = 50 mm by 10 mm
The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN
The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN
By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force
The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa
The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa
The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa
The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN