Answer:
1069.38 gallons
Explanation:
Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.
V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹
Δθ = (5/9)(97°F -52°F) °C = 25 °C.
Let V₂ be its final volume when it cools to 52°F in the tank is
V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)
= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)
= 1.07 × 10³(1 - [0.024]²)
= 1.07 × 10³(1 - 0.000576)
= 1.07 × 10³(0.999424)
= 1069.38 gallons
Answer:
Low Potential energy and High Kinetic energy
Explanation:
Hope this helps and have a good day! Apologies if it's wrong.<3
At the end of the baseball bat, because with the length of the bat he had a longer reach and the end of the bat was moving faster than his hands were
Answer:
The distance by the ball clear the crossbar is 1.15 m
Explanation:
Given that,
Distance = 44 m
Speed = 24 m/s
Angle = 31°
Height = 3.05 m
We need to calculate the horizontal velocity
Using formula of horizontal velocity

Put the value into the formula


We need to calculate the vertical velocity
Using formula of vertical velocity

Put the value into the formula


We need to calculate the time
Using formula of time

Put the value into the formula


We need to calculate the vertical height
Using equation of motion

Put the value into the formula


We need to calculate the distance by the ball clear the crossbar
Using formula for vertical distance

Put the value of h


Hence, The distance by the ball clear the crossbar is 1.15 m
Answer:
x = 17.88[m]
Explanation:
We can find the components of the initial velocity:
![(v_{x})_{o} = 13.3*cos(41.5)=9.96[m/s]\\(v_{y})_{o} = 13.3*sin(41.5)=8.81[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%29_%7Bo%7D%20%20%3D%2013.3%2Acos%2841.5%29%3D9.96%5Bm%2Fs%5D%5C%5C%28v_%7By%7D%29_%7Bo%7D%20%20%3D%2013.3%2Asin%2841.5%29%3D8.81%5Bm%2Fs%5D)
We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.
g = - 9.81[m/s^2]
Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.
![y=y_{o} +(v_{y} )_{o} *t-0.5*g*(t)^{2} \\0=1.9+(8.81*t)-(4.905*t^{2})\\-1.9=8.81*t*(1-0.5567*t)\\t=0\\t=1.796[s]](https://tex.z-dn.net/?f=y%3Dy_%7Bo%7D%20%2B%28v_%7By%7D%20%29_%7Bo%7D%20%2At-0.5%2Ag%2A%28t%29%5E%7B2%7D%20%5C%5C0%3D1.9%2B%288.81%2At%29-%284.905%2At%5E%7B2%7D%29%5C%5C-1.9%3D8.81%2At%2A%281-0.5567%2At%29%5C%5Ct%3D0%5C%5Ct%3D1.796%5Bs%5D)
With this time we can calculate the horizontal distance:
![x=(v_{x})_{o} *t\\x=9.96*1.796\\x=17.88[m]](https://tex.z-dn.net/?f=x%3D%28v_%7Bx%7D%29_%7Bo%7D%20%2At%5C%5Cx%3D9.96%2A1.796%5C%5Cx%3D17.88%5Bm%5D)