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stich3 [128]
3 years ago
6

which has greater cenetic energy a car traveling 30.0 km/hr or one twice as heavy traveling at 15 km/hr?​

Mathematics
1 answer:
iVinArrow [24]3 years ago
8 0

Answer:

30 km/h car

Step-by-step explanation:

From analysis the   car traveling at 30 km/h has greater kinetic energy

we can deduce it from the expression of kinetic energy which is

KE=\frac{1}{2} mv^2

Assuming the mass m= 1 kg

 

For the 30 km/h

KE=\frac{1}{2}*1*30^2 \\\\KE=\frac{1}{2}*1*900\\\\\KE=450 J

   

For the 15 km/h

KE=\frac{1}{2}*2*15^2 \\\\ KE=\frac{1}{2}*2*225 \\\\\ KE=\frac{1}{2}*450 J\\\\\ KE=225 J

Though the kinetic energy is a function of mass and velocity, but from our analysis the faster moving object has more KE

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Find the area, in square feet, of a triangle whose base is 42⁄3 feet and whose altitude is 84⁄7 feet. A. 135⁄21 B. 20 C. 40 D. 2
CaHeK987 [17]

Answer:

Option B. 20\ ft^{2}

Step-by-step explanation:

we know that

The area of a triangle is equal to

A=\frac{1}{2}bh

we have

b=4\frac{2}{3}\ ft

h=8\frac{4}{7}\ ft

Convert mixed numbers to an improper fractions

b=4\frac{2}{3}\ ft=\frac{4*3+2}{3}=\frac{14}{3}\ ft

h=8\frac{4}{7}\ ft=\frac{8*7+4}{7}=\frac{60}{7}\ ft

substitute the values

A=\frac{1}{2}(\frac{14}{3})(\frac{60}{7})\\ \\A=\frac{14*60}{2*3*7}\\ \\A=\frac{840}{42} \\ \\A=20\ ft^{2}

5 0
3 years ago
Read 2 more answers
The cuboid has volume 288cm^3 (a) the cuboid has a length 12 cm and a width 5cm Calculate the height of the cuboid (b) 1cm^3 of
klemol [59]

Answer:

height= 4.8cm

Step-by-step explanation:

volume= length×width× height given the volume and length and width, 288=12×5×h

288=60h

divide both sides by 60

h=4.8cm

7 0
3 years ago
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Given the matrices: 1 2 A= 1 -1 2 1 1 B= 3 4 Calculate AB: C11 C12 [2.1] х 1 2 3 4 C21 C22 C11 = C12 = -2 C22 - C215 DONE​
eduard

Answer:

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

Step-by-step explanation:

Given the matrices

A=\begin{pmatrix}1&-1\\ 2&1\end{pmatrix}

B=\begin{pmatrix}1&2\\ \:3&4\end{pmatrix}

Calculating AB:

\begin{pmatrix}1&-1\\ \:\:2&1\end{pmatrix}\times \:\begin{pmatrix}1&2\\ \:\:3&4\end{pmatrix}=\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}

Multiply the rows of the first matrix by the columns of the second matrix

                                   =\begin{pmatrix}1\cdot \:1+\left(-1\right)\cdot \:3&1\cdot \:2+\left(-1\right)\cdot \:4\\ 2\cdot \:1+1\cdot \:3&2\cdot \:2+1\cdot \:4\end{pmatrix}

                                   =\begin{pmatrix}-2&-2\\ 5&8\end{pmatrix}

Hence,

\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}=\begin{pmatrix}-2&-2\\ \:5&8\end{pmatrix}

Therefore,

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

5 0
3 years ago
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Find the value of w in the pentagon below if the perimeter is 105 meters.
gavmur [86]

Answer: 9 (with a margin of: 0)

Step-by-step explanation:

Times everything by 5

That gives you 15 + 10w = 105.

Subtract 15 from both sides

Divide by 10 to get w by itself

Then you solve

5 0
3 years ago
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AlexFokin [52]
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3 years ago
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