6. A golf ball is hit a distance of 300 yards in 10 sec. What is the speed of the golf ballo
1 answer:
The speed of the ball is 27.4 m/s
Explanation:
The speed of an object is given by:
where
d is the distance covered
t is the time taken
In this problem, we have:
d = 300 yards is the distance covered by the golf ball
t = 10 s is the time taken
Keeping in mind that
1 yard = 0.914 m
We can convert the distance from yards to meters:
And substituting into the equation, we find the speed of the ball:
Learn more about speed:
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Answer:
<em>The coefficient of static friction between the crate and the floor is 0.41</em>
Explanation:
<u>Friction Force</u>
When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.
The friction force when an object is moving on a horizontal surface is calculated by:
[1]
Where is the coefficient of static or kinetics friction and N is the normal force.
If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:
N = W = m.g
The crate of m=20 Kg has a weight of:
W = 20*9.8
W = 196 N
The normal force is also N=196 N
We can find the coefficient of static friction by solving [1] for :
The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:
The coefficient of static friction between the crate and the floor is 0.41
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s