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
I assume that the force of 20 N is applied along the direction of motion and was applied for the whole 6 meters, the formula of work is this; Work = force * distance * cosθ where θ is zero degrees. Plugging in the data to the formula; Work = 20 N * 6 m * cos 0º.
Work = 20 N * 6 m * 1
Work = 120 Nm
Work = 120 joules
Hope this helps!
<h3>Answer:</h3>
The mechanical advantage would decrease, making the block more difficult to lift.
<h3>Explanation:</h3>
The mechanical advantage in such a setup is the ratio of distance from A to B to the distance from D to B. In this picture, that ratio is less than 1, meaning the advantage of having this setup is less than the advantage of no setup at all.
While the force required to lift the block is increased by this setup, the distance over which that force is applied will be smaller for raising the block to a given height. (Overall, for the same height, more work is required with the lever setup because you're raising part of the mass of the lever as well as the mass of the block.)
likely due to the location.
<span>Three-fourths of the illuminated side of the moon be visible
from Earth when the moon is at position-B. (choice-B)
Notice the particularly egregious screw-up on the drawing at the
"Third Quarter" position. There, the drawing shows the illuminated
side of the moon AWAY from the sun, and the dark side of the moon
TOWARD the sun. That's about as silly as you can get. </span>
