2.00 A
they want me to put an explanation but im not bout to do that uhnuhn
Answer: 1.81m
Explanation:
Given the following:
Horizontal speed = 4m/s
Height of table = 1m
Taking the vertical component of the motion:
Using:
S = ut + 0.5 at^2
Where, S = height of table, a = acceleration due to gravity (9.8m/s^2), t = time and u = Initial Velocity in vertical direction = 0
1 = 0(t) + 0.5(9.8)t^2
1 = 4.9t^2
t^2 = 1 / 4.9
t^2 = 0.204081
t = sqrt(0.204081)
t = 0.452s
For Horizontal :
Gravitational acceleration is absent, therefore distance can be calculated using:
Distance = Velocity × time
Distance = 4m/s × 0.452s
Distance = 1.808 m
Distance = 1.81m
A golfer hits a 45 g golf ball during 2.0 × 10⁻³ s causing its final speed to be 38 m/s. The impulse of the golf ball is 1.7 kg.m/s. The average force that the club exerts on the golf ball is 850 N and the average force that the golf ball exerts on the club is -850 N.
A golfer hits a 45 g (m) golf ball resting on a tee so that the golf ball leaves the tee at the horizontal speed of 38 m/s (v). To answer the questions, we need to consider the concepts of impulse and linear momentum.
Impulse (I) equals the average net external force (F) multiplied by the time (t).
I = F × t [1]
Linear momentum (p) is defined as the product of a system's mass (m) multiplied by its velocity (v).
p = m × v [2]
The impulse experienced by the object equals the change in the linear momentum of the object.
I = F × t = m × Δv [3]
We will use the equation [3], considering that Δv = v because it starts from the rest.
I = m × v = 0.045 kg × 38 m/s = 1.7 kg.m/s
The club and the golf ball are in contact for 2.0 × 10⁻³ s (t). We will calculate the average force that the club exerts on the golf ball (Fcg) using the equation [1].
I = Fcg × t
Fcg = I / t = (1.7 kg.m/s)/(2.0 × 10⁻³ s) = 850 N
According to Newton's third law of motion, action and reaction have the same value and opposite signs. Thus, the average force that the golf ball exerts on the club (Fgc) is -850 N.
A golfer hits a 45 g golf ball during 2.0 × 10⁻³ s causing its final speed to be 38 m/s. The impulse of the golf ball is 1.7 kg.m/s. The average force that the club exerts on the golf ball is 850 N and the average force that the golf ball exerts on the club is -850 N.
Learn more about impulse here: brainly.com/question/904448
Answer:
Correct answer: (1) I = 1.71 kg m/s, (2) F = 855 N
Explanation:
Given:
The mass of the ball m = 45 g = 45 · 10⁻³ kg
Initial velocity V = 38 m/s
Contact time t = 2 · 10⁻³ s
(1) I = ?
The impulse is calculated according to the formula:
I = m · V = 45 · 10⁻³ · 38 = 1,710 · 10⁻³ = 1.71 kg m/s
I = 1.71 kg m/s
(2) F = ?
The average force is calculated according to the formula:
F = ΔI / Δt = 1.71 / (2 · 10⁻³) = 855 N
F = ΔI / Δt = 1.71 / (2 · 10⁻³) = 855 N
F = 855 N
God is with you!!!