Answer:
Vertical distance= 3.3803ft
Explanation:
First with the speed of the ball and the distance traveled horizontally we can determine the flight time to reach the plate:
Velocity= (90 mi/h) × (1 mile/5280ft) = 475200ft/h
Distance= Velocity × time⇒ time= 60.5ft / (475200ft/h) = 0.00012731h
time= 0.00012731h × (3600s/h)= 0.458316s
With this time we can determine the distance traveled vertically taking into account that its initial vertical velocity is zero and its acceleration is that of gravity, 9.81m/s²:
Vertical distance= (1/2) × 9.81 (m/s²) × (0.458316s)²=1.0303m
Vertical distance= 1.0303m × (1ft/0.3048m) = 3.3803ft
This is the vertical distance traveled by the ball from the time it is thrown by the pitcher until it reaches the plate, regardless of air resistance.
Verrrrry interesting question !
Force= (mass) x (acceleration)
(49.5kg) x (1.3 m/s²) =
<u> Net</u> force on the box is <u>64.25 newtons upward</u>.
That's the <u>net</u> force. It's the combination of the tension in the rope
pulling the box up, and the force of gravity pulling the box down.
Force of gravity = "weight" = (mass) x (gravity) =
(49.5) x (9.8) = <u>485.1 newtons downward</u>
The tension in the rope must be enough to balance the downward force
of gravity PLUS provide another 64.25 newtons upward.
<em>Tension =</em> (485.1) + (64.25) = <em>549.35 newtons</em>
The answer is the third choice, "the brightness of the beam of light increases"
According to Einstein’s theory, an increase in the number of photons (per unit are) affects a beam of light by causing a higher intensity. <span> Sometimes, the term "brightness" is used when referring to the </span>intensity<span> of a color, a</span><span>lthough there are instances where this can be a misleading term when we try to describe </span><span>intensity</span>
a) Hooke's law:
F = kΔx
F = spring force, k = spring constant, Δx = change of spring length
Given values:
F = 30N, Δx = 0.1m
Plug in and solve for k:
30 = k(0.1)
k = 300N/m
b) Apply the work-energy theorem; whatever work you put into deforming the spring becomes stored as spring potential energy:
W = 0.5kΔx²
W = work, k = spring constant, Δx = change of spring length
Given values:
k = 300N/m (from part a), Δx = 0.5m
Plug in and solve for W:
W = 0.5(300)(0.5)²
W = 37.5J
c) Apply the work-energy theorem here:
W = 0.5kΔx²
Given values:
k = 300N/m (from part a), Δx = 0.4m
Plug in and solve for W:
W = 0.5(300)(0.4)²
W = 24J
d) To find how much additional work needs to be done to stretch the spring an additional 0.1m if it's already been stretched 0.1m, find the potential energies for when the spring is stretched 0.2m and 0.1m and subtract them:
W = 0.5kΔx₂² - 0.5kΔx₁² = 0.5k(Δx₂² - Δx₁²)
Given values:
k = 300N/m, Δx₂ = 0.2m, Δx₁ = 0.1m
Plug in and solve for W:
W = 0.5(300)(0.2² - 0.1²)
W = 4.5J
Answer:
Vf=140 m/s
Vi=0
t=50s
Explanation:
1. Acceleration is called change of velocity.
2. a=Vf-Vi/t
a=140-0/50
a=140/50
a=2.8m/s^2
