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2 years ago

6

the acceleration due to gravity on the moon is 1.6 m/s^2 what is the gravitational potential energy of a 1200 kg lander resting

on top of a 350m hill

1 answer:

Mumz [18]2 years ago

5 0

<u>Answer</u>

672,000 Joules

<u>Explanation</u>

Gravitational potential energy (P.E) is the energy possessed by a body that is at a potential height from the ground.

IT is calculated by the formula;

P.E = mgh

Where m ⇒ mass

g ⇒ acceleration due to gravity

h ⇒ height from trhe ground.

P.E = 1200 × 1.6 × 350

= 672,000 Joules.

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It requires 0.30 kJ of work to fully drive a stake into the ground. If the average resistive force on the stake by the ground is

LenaWriter [7]

Answer:

Length of the stake will be 0.3623 m

Explanation:

We have given energy required to fully drive a stake into ground = 0.30 KJ = 300 J

Average resistive force acting on the floor is equal to F = 828 N

We have to find the length of the stake

We know that work done is given by

W = Fd, here W is work done , F is average force and d is the length of the stake

So 300 = 828×d

d = 0.3623 m

So length of the stake will be 0.3623 m

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2 years ago

A car is pulled with a force of 10,000 N. The car's mass is 1267 kg. But, the car covers 394.6 m in 15 seconds

solong [7]

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/ $s^{2}$

B. Formula: a= $\frac{V-V_{0} }{t}$ and s=d/t

speed= 394.6/15

s=26.3m/s

a= $\frac{26.3-0}{15}$

a=1.75m/ $s^{2}$

C. 7.9-1.75=difference of 6.15m/ $s^{2}$

D. The force that most likely caused this difference is friction forces

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1 year ago

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100-ft-long horizontal pipeline transporting benzene develops a leak 43 ft from the high-pressure end. The diameter of the leak

Amanda [17]

Answer:

Explanation:

The mass flow rate of benzene from the leak in the pipeline containing benzene is:

$Q_m=AC_o\sqrt{2\rho g_cP_g}$

Here, $Q_m$ is the mass flow rate through the leak of the pipeline. $A$ is the area of the hole, $C_o$ is the discharge rate, $\rho$ is the fluid density, $g_c$ is the gravitational constant and $P_g$ is the constant gauge pressure within the process unit.

The diametre of the leak (d) is 0.1 in. Convert from in to ft.

$d=(0.1 in)(\frac{1ft}{12in})\\=8.33\times 10^{-3}ft$

Calculate the area (A) of the hole. The area of the hole is.

$A=\frac{\pi d^2}{4}$

Substitute 3.14 for $\pi$ and $8.33\times 10^{-3}ft$ for d and calculate A.

$A=\frac{\pi d^2}{4}\\\\\frac{(3.14)(8.33\times 10^{-3})^2}{4}\\\\5.45\times 10^{-5}ft^2$

The specific gravity of benzene is 0.8794. Specific gravity is the ratio of th density of a substance to the density of a reference substance.

Specific gravity of benzene = density of benzenee/denity of reference substance

Rewrite the expression in terms of density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

Take the reference substance as water. Density of water is $62.4\frac{Ib_m}{ft^3}$ . Calculate density of benzene.

Density of benzene = specific gravity of benzene x density of reference substance

$=(0.8794)(62.4\frac{Ib_m}{ft^3})\\\\54.9\frac{Ib_m}{ft^3}$

Calculate the pressure at the point of leak. The pressure is the average of the pressure of the high and low pressure end. Write the expression to calculate the average pressure.

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

Calculate the distance from the downstream pressure end. The distance from upstream pressure end is 43 ft. Total of the pipe is 100 ft.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

The distance from upstream pressure end is 43 ft. Total length of the pipe is 100 ft. Substitute the values in the equation.

Distance from the downstream pressure end = Total length of the pipe - Distance from the upstream pressure end

= 100ft - 43ft = 57 ft

Substitute 50 psig for upstream, 43 ft fr distance from the upstream pressure end, 40 psig for downstream pressure, 57 ft for distance from the downstream pressure end, and 100 ft for the total length of the horizontal pipeline and calculate $P_g$ .

Upstream x distance from upstream pressure end

$P_g$ =+DOWNSTREAM PRESSURE X DISTANCE FROM THE DOWNSTREAM PRESSURE END/ TOTAL LENGTH OF THE HORIZONTAL PIPELINE

$=\frac{(50psig\times 43ft)+(40psig \times 57ft)}{100ft}\\\\=44.3psig$

Convert the pressure from psig to $Ib_f/ft^2$

$P_g=(44.3psig)(\frac{1\frac{Ib_f}{ft^2}}{1psig})(144\frac{in^2}{ft^2})\\\\=6,379.2\frac{Ib_f}{ft^2}$

The leak is like a sharp orifice. Take the value of the discharge coefficient as 0.61.

Substitute $5.45\times 10^{-5}ft^2$ for A. 0.61 for $C_o$ , $54.9\frac{Ib_m}{ft^3}$ for $\rho$ , $32.17\frac{ft.Ib_m}{Ib_f.s^2}$ for $g_c$ , and $6,379.2\frac{Ib_f}{ft^2}$ for $P_g$ and calculate $Q_m$

$Q_m=AC_o\sqrt{2\rho g_cP_g}\\\\=(5.45\times 10^{-5}ft^2)(0.61)\sqrt{2(54.9\frac{Ib_m}{ft^3})(32.17\frac{ft.Ib_m}{Ib_f.s^2})(6,379.2\frac{Ib_f}{ft^2})}\\\\(3.3245\times 10^{-5}ft^2)\sqrt{22,533,031.21\frac{Ib^2_m}{ft^4.s^2}}\\\\=0.158\frac{Ib_m}{s}$

The mass flow rate of benzene through the leak in the pipeline is $0.158\frac{Ib_m}{s}$

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2 years ago

The charges for four ions are sodium (Na) +1, calcium (Ca) +2, fluorine (F) –1, and sulfur (S) –2. Which would be an ionic bond

Ludmilka [50]

The correct answer would be B. It is the compound Na2S that forms an ionic bond from the given ion above. From the choices, it is the only compound that exist and is available. Metals cannot react with another metal as well as nonmetal to nonmetal.

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2 years ago

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Cylinder A is moving downward with a velocity of 3 m/s when the brake is suddenly applied to the drum. Knowing that the cylinder

Xelga [282]

Answer:

Incomplete question

Check attachment for the given diagram

Explanation:

Given that,

Initial Velocity of drum

u=3m/s

Distance travelled before coming to rest is 6m

Since it comes to rest, then, the final velocity is 0m/s

v=3m/s

Using equation of motion to calculate the linear acceleration or tangential acceleration

v²=u²+2as

0²=3²+2×a×6

0=9+12a

12a=-9

Then, a=-9/12

a=-0.75m/s²

The negative sign shows that the cylinder is decelerating.

Then, a=0.75m/s²

So, using the relationship between linear acceleration and angular acceleration.

a=αr

Where

a is linear acceleration

α is angular acceleration

And r is radius

α=a/r

From the diagram r=250mm=0.25m

Then,

α=0.75/0.25

α =3rad/sec²

The angular acceleration is =3rad/s²

b. Time take to come to rest

Using equation of motion

v=u+at

0=3-0.75t

0.75t=3

Then, t=3/0.75

t=4 secs

The time take to come to rest is 4s

7 0

1 year ago