All categories

3 years ago

An earthquake causes a 3 kg book to fall from a shelf. If the book lands with a speed of 4.0 m/s, from what height did it fall?

1 answer:

6 0

Given:
3kg book
speed of falling of 4.0 m/s
acceleration due to gravity -9.8 m/s²

Vf² = Vi² + 2 * a * d

Vf = final velocity ; Vi = initial velocity ; a = acceleration ; d = distance

0² = (4 m/s)² + 2(-9.8m/s²) * d
0 = 16 m²/s² -19.6 m/s² * d
-16 m²/s² = -19.6 m/s² * d
-16 m²/s² / -19.6 m/s² = d
0.82 m = distance

The 3 kg book fell from the a shelf at the height of 0.82 meters.

You might be interested in

A regular hexagon has sides of 3 feet. What is the area of the hexagon?

Dominik [7]

Answer:

13.5√(3) ft^2

Step-by-step explanation:

In this, question, we are to calculate the area of the hexagon

Mathematically, we can calculate the area of a hexagon using the formula;

A = {3√(3)/2}a^2

where a is the length of the side which is 3 feet

substituting the value of 3 into the formula, we have;

A = 3√3/2 * 3^2 = 27 √(3)/2 = 13.5√(3) ft^2

7 0

3 years ago

Solve each equation algebraically using if–then statements to justify each step 2(x + 4) = 8

Alex

This is what Photomath got

8 0

3 years ago

Lilly bought stock at $83.60. The next day, the price increased by $15.35. Then this price changed by $-2.15.Wgat was the final

rodikova [14]

The final stock price would be $96.8

3 0

3 years ago

a pyramid whose altitude is 5ft weighs 800lbs. at what distance from its vertex must it be cut by a plane parallel to its base s

slavikrds [6]

A structure which has a square base and four triangular sides meeting at a point is called pyramid.

At distance of 3.97 feet from its vertex , pyramid is cut by plane so that the two solids of equal weight will be formed.

It is assumed that weight of pyramid is proportional to its volume.

So, $w = k V$ , where V is volume of original pyramid and v is volume of small pyramid and k is constant.

Let us consider that at h distance, pyramid is cut from its vertex. So a small pyramid is also formed.

Assume that base area of original pyramid is A and base of small pyramid is a.

Volume of original pyramid is, $V= \frac{1}{3} *A* 5$

So, weight of original pyramid, $W = k *(\frac{1}{3} *A* 5) = 800$

Volume of small pyramid is, $v = \frac{1}{3}* a* h$

So, weight of small pyramid, $w = k*(\frac{1}{3}* a* h)=400$

Since, base and height of small pyramid and original pyramid are in proportion.

So, $\frac{a}{A} = (\frac{h}{5}) ^{2}$

$a = (\frac{h}{5} )^{2}A$

Substituting value of a in equation $k*(\frac{1}{3}* a* h)=400$

So, $k*(\frac{1}{3}* \frac{h^{2} }{25}A * h)=400$

$(k*\frac{1}{3}* A*5)*\frac{1}{5} *\frac{h^{2} }{25} * h)=400$

Since, $(k *\frac{1}{3} *A* 5) = 800$ , substitute in above equation.

So, $800*\frac{h^{3} }{125}=400\\\\h^{3}=\frac{125}{2}\\\\h=\sqrt[\frac{1}{3} ]{62.5}\\\\h=3.97 feet$

Learn more:

brainly.com/question/17950304

6 0

2 years ago

Read 2 more answers

Name the intersection of plane QMW and plane RMW

Norma-Jean [14]

The answer is "C", "MW".

In the given problem, the place QMW and plane RMW. These planes intersect at MW, in which intersection is either a point, line or curve that an entity or entities both possess or is in contact with but if we see in Euclidean geometry, the intersection of two planes is called a “line”. In the plane we can understand that the common line for both plane QMW and plane RMW is MW.

6 0

3 years ago

Read 2 more answers