Explanation:
Given that,
Net force = 135.5 N
Mass = 26.7 kg
(a). We need to draw a free body diagram
A force is exerted on a object.
(b). We need to calculate the acceleration
Using formula of acceleration
(c). We need to draw the sketch a position vs. time graph for the object.
Using equation of motion
Put the value into the formula
(d). We need to draw the sketch a velocity vs. time graph for the object.
Using equation of motion
Put the value into the formula
(e). We need to calculate the distance travel in 8.82 s
Using equation of motion
Hence, This is the required solution.
Answer:
P is pressure ,rho(slanted p) is density,g is the acceleration due to gravity, v is velocity and h is height
Explanation:
paraan ito ng pagrerespeto
The answer is: " 208 g " .
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Explanation:
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The formula/ equation for density is:
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D = m / V ; That is, "mass divided by volume" ;
Density is expressed as:
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"mass per unit volume"; in which the "mass" is expressed in units of "g" ("grams") ; and the "unit volume" is expressed in units of:
"cm³ " or "mL";
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{Note the exact equivalent: 1 cm³ = 1 mL }.
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→ The formula is: " D = m / V " ;
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in which:
"D" refers to the "density" (see above), which is: "8.9 g/cm³ " (given);
"m" refers to the "mass" , in units of "g" (grams), which is unknown; and we want to find this value;
"V" refers to the "volume", in units of "cm³ " ;
which is: "23.4 cm³ " (given);
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We want to find the mass, "m" ; so we take the original equation/formula for the density:
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D = m / V ;
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And we rearrange; to isolate "m" (mass) on ONE side of the equation; and then we plug in our known/given values;
to solve for "m" (mass); in units of "g" (grams) ;
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Multiply each side of the equation by "V" ;
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V * { D = m / V } ; to get:
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V * D = m ; ↔ m = V * D ;
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Now, we plug in the given values for "V" (volume) and "D" (density) ; to solve for the mass, "m" ;
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m = V * D ;
m = (23.4 cm³) * (8.9 g / 1 cm³) = (23.4 * 8.9) g = 208.26 g ;
→ Round to "208 g" (3 significant figures);
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The answer is: " 208 g " .
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Answer:
B) 18,000 feet MSL
Explanation:
There are three-dimensional parts in the navigation airspace in the world. The class E airspace is mostly used in the regions with coastal areas that are relatively populated. If we consider certain forms of exceptions, the class E airspace can move in the upward direction to few feet (i.e. 1200 ft). However, this doesn't include 18,000 feet MSL.
