Answer: M.A = 3
Explanation:
A ramp is an example of an inclined plain. Where the
Height H = 1.5 m
Length L = 4.5 m
Mechanical advantage of a machine is the ratio of the load to effort. While mechanical advantage M.A of an inclined plain is the ratio of the length of the plain to the height of the plain.
M.A = L/H
Substitute the values of L and H into the formula
M.A = 4.5/1.5 = 3
The mechanical advantage of the ramp is 3
Answer:
100 meters, 54.5 East of North or 125.5 North of East.
Explanation:
Try drawing it out to get a better visual. Make sure that when you draw the arrows that you make a scale (for example: 1 cm = 10 meters). After drawing it out, draw a line from the origin/starting point and connect it to the end point from the "75 m west" arrow. Then, measure the line you drew and convert it back into meters. Lastly, measure the angle.
Answer: SG = 2.67
Specific gravity of the sand is 2.67
Explanation:
Specific gravity = density of material/density of water
Given;
Mass of sand m = 100g
Volume of sand = volume of water displaced
Vs = 537.5cm^3 - 500 cm^3
Vs = 37.5cm^3
Density of sand = m/Vs = 100g/37.5 cm^3
Ds = 2.67g/cm^3
Density of water Dw = 1.00 g/cm^3
Therefore, the specific gravity of sand is
SG = Ds/Dw
SG = (2.67g/cm^3)/(1.00g/cm^3)
SG = 2.67
Specific gravity of the sand is 2.67
<h2><u>Answer:</u></h2>
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:
1. (There would be variables that would be hard to control, leading to less reliable data.)
3. (Meter sticks may lack precision or may be read incorrectly.)
4. (Real glider data may vary since real collisions may involve loss of energy.)
5. (Human error in recording or plotting the data could be a factor.)
Answer:
60.8 cm²
Explanation:
The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².
σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²
Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.
Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface
So, q = ε₀Ф = σA'
ε₀Ф = σA'
making A' the area of the Gaussian surface the subject of the formula, we have
A' = ε₀Ф/σ
A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²
A' = 81.4568/1.34 × 10⁻⁴ m²
A' = 60.79 × 10⁻⁴ m²
A' ≅ 60.8 cm²
