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Ivahew [28]
2 years ago
10

You are wallpapering two walls of a room. One wall measures 15 ft by 12 ft and the other measures 9 ft by 12 ft. The wall paper

costs $19.40 per roll, and each roll covers 50 square feet. What does it cost to cover the two walls?
Physics
1 answer:
enyata [817]2 years ago
4 0

Answer:

5.76 round off to 6

Explanation:

wall 1 = 15 × 12 = 180

wall 2 = 9 × 12 = 108

now 1 roll covers 50 square feet

formula = wall 1 + wall 2 / 50

= 180 + 108 / 50

= 288÷ 50

= 5.76

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Wile E. Coyote wants to launch Roadrunner into the air using a long lever asshown below. The lever starts at rest before the Coy
algol13

Answer:

Explanation:

Find attached the solution

4 0
2 years ago
A crane uses a block and tackle to lift a 2200N flagstone to a height of 25m
Cloud [144]

Remember the headline:  ENERGY IS NEVER CREATED OR DESTROYED

The amount of energy before and after are always equal.  All we ever do with energy is move it around from one place to another.

a). A crane can't create energy.  Lifting the same rock in 20 different ways always takes the <u><em>same amount of work</em></u>.  It doesn't matter whether one person picks the rock straight up, or 50 people get around it and lift it, or roll it up a ramp, or lift it with 16 pulleys and a mile of rope, or use a giant steam crane.

You want to lift a 2200N weight up 25m, you're going to have to supply

(2200N) x (25m) = <em>55,000 Joules</em> of work.

c). YOU put out 55,000 Joules of energy.  It had to GO someplace. Where is it now ? ===>  It's the potential energy the rock has now, from being 25m higher than it was before.  That <em>55,000 Joules</em> is NOW the potential energy  of the rock.

No energy was created or destroyed.  It just got moved around.  

55,000 Joules of energy began as nuclear energy in the core of the sun. Solar radiation carried it to the Earth. Plants absorbed it, and stored it as chemical energy.  You ... or a cow that you ate later ... ate the plants and took the chemical energy.  One way or the other, the chemical energy got stored in your blood and fat.  When you needed to put it out somewhere, you moved it into your muscles, and they converted it into mechanical energy.  Then you used the mechanical energy to exert forces.  Today, you used the original 55,000 joules to lift the flagstone, and NOW that energy is in the flagstone, 25 meters up off the ground !

6 0
2 years ago
James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m
Over [174]

Answer:

4.1\cdot 10^8 N

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

p=p_0 + \rho g h

where

p_0 =1.01\cdot 10^5 Pa is the atmospheric pressure

\rho = 1000 kg/m^3 is the water density

g=9.8 m/s^2 is the gravitational acceleration

h=11,000 m is the depth

Substituting,

p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2

And so, the inward force exerted on it is

F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N

8 0
2 years ago
Read 2 more answers
Find the object's speeds v1, v2, and v3 at times t1=2.0s, t2=4.0s, and t3=13s.
Burka [1]
Since this is a distance/time graph, the speed at any time is the slope
of the part of the graph that's directly over that time on the x-axis.

At time  t1 = 2.0 s
That's in the middle of the first segment of the graph,
that extends from zero to 3 seconds.
Its slope is  7/3 .              v1 = 7/3 m/s .

At time  t2 = 4.0 s
That's in the middle of the horizontal part of the graph
that runs from 3 to 6 seconds.
Its slope is zero.
                                     v2 = zero .

At time  t3 = 13 s.
That's in the middle of the part of the graph that's sloping down,
between 11 and 16 seconds.
Its slope is  -3/5 .            v3 = -0.6 m/s .              
7 0
2 years ago
Read 2 more answers
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

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(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
2 years ago
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