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juin [17]
3 years ago
7

The time T in seconds for a pendulum of length L feet to make one swing is given by Upper T=2\pi \sqrt((L)/(36)). How long is a

pendulum (to nearest hundredth) if it makes one swing in 2.1 seconds? Use 3.14 for \pi .
Physics
1 answer:
Andreyy893 years ago
5 0

Answer:

3.6ft

Explanation:

Using= 2*π*sqrt(L/32)

To solve for L, first move 2*n over:

T/(2*π) = sqrt(L/32)

Next,eliminate the square root by squaring both sides

(T/(2*π))2 = L/32

or

T2/(4π2) = L/32

Lastly, multiply both sides by 32 to yield:

32T2/(4π2) = L

and simplify:

8T²/π²= L

Hence, L(T) = 8T²/π²

But T = 2.1

Pi= 3.14

8(2.1)²/3.14²

35.28/9.85

= 3.6feet

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in a lever, a load of 600N is lifted by using 400 effort. If the load is at the distance of 20cm and the effort at the distance
koban [17]

Explanation:

Load (l) = 680N

Effort (E) = 500N

Length slope (l) = 12m

Height slope (h) = 8 m

Output = load * height

680 *8 = 5.44 *103 J

The Input = effort * length = 500 *12 = 6000J

the Mechanical advantage (M.A) = load effort= 600500=1.36

the Velocity ratio (V.R) =lh=128 = 1.5

the Efficiency =M.A100%V.R= 90.6%

8 0
3 years ago
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supp
IRINA_888 [86]

Complete question :

NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?

Answer:

601000 N

Explanation:

Given that :

Acceleration due to gravity at lunar outpost = 1.6m/s²

Supported Weight of supplies = 1 * 10^5 N

Acceleration due to gravity on the earth surface = 9.8m/s²

Maximum weight of supplies as measured on EARTH :

Ratio of earth gravity to lunar post gravity:

(Earth gravity / Lunar post gravity) ;

(9.8 / 1.63) = 6.01

Hence, maximum weight of supplies as measured on EARTH should be :

6.01 * (1 × 10^5)

6.01 × 10^5

= 601000 N

3 0
3 years ago
Travels 11,000 feet along a dark desert highway if the car averages 84 mph find the amount of time to cover this distance
laila [671]

The time taken by traveler to cover the distance is,

t=\frac{d}{v}

Substitute the known values,

\begin{gathered} t=\frac{(11000\text{ ft)}}{(84\text{ mph)(}\frac{1.46667\text{ ft/s}}{1\text{ mph}})_{}} \\ \approx89.3\text{ s} \end{gathered}

Therefore, the time taken by traveler to cover the distance is 89.3 s.

5 0
1 year ago
Calculating Net Force
VARVARA [1.3K]

Explanation:

225=m (2.20m/s2 which give m=16kg I used newtons second law to fins the required average force

6 0
3 years ago
] A new coal-fired 750 MWe power plant with a thermal efficiency of 42% burns 9000 Btu/lb coal, which contains 1.1% sulfur. a. I
Valentin [98]

Answer:

The net emissions rate of sulfur is 1861 lb/hr

Explanation:

Given that:

The power or the power plant = 750 MWe

Since the power plant with a thermal efficiency of 42% (i.e. 0.42) burns 9000 Btu/lb coal, Then the energy released per one lb of the coal can be computed as:

\mathtt{=( 0.42\times 9000\times 1055.06) J}

= 3988126.8  J

= 3.99 MJ

Also, The mass of the burned coal per sec can be calculated by dividing the molecular weight of the power plant by the energy released per one lb.

i.e.

The mass of the coal that is burned per sec =\dfrac{750}{3.99}

The mass of the coal that is burned per sec = 187.97 lb/s

The mass of sulfur burned  = \dfrac{1.1}{100} \times 187.97 \  lb/s

= 2.067 lb/s

To hour; we have:

= 7444 lb/hr

However, If a scrubber with 75% removal efficiency is utilized,

Then; the net emissions rate of sulfur is (1 - 0.75) × 7444 lb/hr

= 0.25 × 7444 lb/hr

= 1861 lb/hr

Hence, the net emissions rate of sulfur is 1861 lb/hr

8 0
3 years ago
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