For the answer to the question above, first find out the gradient.
<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>
<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>
<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>
<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>
<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
L = r x p = rmv = mr²ω
L = 0.25 x 0.75² x 12.5 = 1.758
Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²
2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads
3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²
4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J
5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²
6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²
7. v = ω r = 28rad/s * ½(0.33m) = 4.62 m/s
8. s = Θ r = 266 rads * 0.33m = 88 m
Answer:
The correct answer is -
A (the entire green box): Chemical Equation
B (the blue box): Reactants
C (the arrow): Reacts to Form
D (the number): Coefficient
E (the purple box): Products
Explanation:
The chemical reaction of burning methane and oxygen is as follows;
Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.
2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products
Answer:When the length of a string is changed, it will vibrate with a different frequency.Shorter strings have higher frequency and therefore higher pitch.