Ask question

Ask question

All categories

3 years ago

14

A rectangular walkway is 5 times as long as it is wide. The perimeter of the walkway is 360 inches. What is the area, in square

inches, of the walkway

1 answer:

Ksenya-84 [330]3 years ago

3 0

Answer:

Area of a rectangle, $A=4500\ inch^2$

Explanation:

Let, the length of the rectangle is l and width is b. It is given that, a rectangular walkway is 5 times as long as it is wide.

According to question, l = 5b.....(1)

The perimeter of the rectangle is 360 inches. It is given by:

$P=2(l+b)$

$360=2(l+5b)$

b = 30 inches

Length, l = 150 inches (from equation 1)

Area of a rectangle is given by :

$A=l\times b$

$A=30\times 150$

$A=4500\ inch^2$

Hence, this is the required solution.

You might be interested in

Which is the correct reasoning for why a gamma ray telescope located in Antarctica that is to be used to look for evidence of bl

steposvetlana [31]

Answer:

Well, the options are not given in the questions. But the correct option is:

"Gamma rays do not completely penetrate earth atmosphere."

Explanation:

Most of the gamma rays coming from the stars and sun can not penetrate the earth's atmosphere because atmosphere is as thick to the gamma rays as a 12 foot thick plate of aluminium. They rather absorb in the atmosphere and knock out electrons by reacting with the atoms and produce ions and electrons. These secondary ions and electrons then came to earth during rain fall. Due to this reason gamma ray telescope was not get funded for the study of black holes.

6 0

4 years ago

Read 2 more answers

An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and ind

torisob [31]

Answer:

This satisfy the above given condition so we can say that this capacitor.

Explanation:

Let's take one by one option and check whether is wrong or right

<u>For inductor:</u>

$I=I_osin(wt-\frac{\pi }{2})$

Given that at t=T/4 ,I=0 and we know that

$w=\dfrac{2\pi }{T}$

So at T/4

$I=I_osin(\frac{2\pi }{T}\times \frac{T}{4}-\frac{\pi }{2})$

I=0 A

At t=T/2

$I=I_osin(\frac{2\pi }{T}\times \frac{T}{2}-\frac{\pi }{2})$

$I=I_o$

It means that this not a indutor.

<u>For capacitor:</u>

$I=I_osin(wt+\frac{\pi }{2})$

At T/4, I=0

At t=T/2

$I=I_osin(\frac{2\pi }{T}\times \frac{T}{2}+\frac{\pi }{2})$

$I= -I_o$

This satisfy the above given condition so we can say that this capacitor.

6 0

4 years ago

A good example of kinetic energy would be_____.

cestrela7 [59]

Answer:

B.) a stretched rubber band

Explanation:

-

4 0

3 years ago

An unbanked circular highway curve on level ground makes aturn

tino4ka555 [31]

Answer:

Explanation:

Given

Velocity of traffic $v=60\ mi/hr\approx 26.82\ m/s$

Maximum value of Centripetal force is one-tenth of weight

such that $(F_c)_{max}=\frac{mg}{10}$

to make a safe turn centripetal force with radius of curvature r is given by

$F_c=\frac{mv^2}{r}$

$(F_c)_{max}=\frac{mg}{10}=F_c=\frac{mv^2}{r}$

$\frac{mg}{10}=\frac{mv^2}{r}$

$r=\frac{10v^2}{g}$

$r=\frac{10\times 26.82^2}{9.8}$

$r=733.99\approx 734\ m$

4 0

4 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s lat

victus00 [196]

Answer / Explanation

It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.

So for proper understanding, the two parts of the question will be highlighted.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?

b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.

Solve for Vmax

Step Process

a) Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]

Where V₀ = 6m/s,

We have,

h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]

= 0.411 m

Where V₀ = 9.5m/s

We have,

h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]

= 1152 m

b) From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.

Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.

Therefore Vₙ = V₀max = 9.8 m/s

4 0

3 years ago