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3 years ago

DUE TONIGHT HURRY PLS I give 30 pts

Physics

1 answer:

Paha777 [63]3 years ago

6 0

The answer is both.

For kinetic energy:

KE = 1/2*m*v^2 = 0.5*20,000 grams*5 = 50,000 J

For gravitational potential energy:

Pe = mgh = 20,000 grams*9.81 m/s^2*2 m = 392.2 J

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If each washer has a mass of 4.9 g, then the mass of two washers in kilograms would be Given that F = ma, and if 4 washers are a

zysi [14]

Given that,

Mass of each washer = 4.9 g

We need to calculate the mass of two washers in kg

Using conversion of unit

Mass of each washer $m= 4.9\times10^{-3}\ Kg$

So, Mass of two washers is

$M=2\times m$

Put the value of m

$M=2\times4.9\times10^{-3}\ k$

$M=0.0098\ Kg$

If 4 washer are attached to the spring

We need to calculate the applied force on the car

Using formula of force

$F=mg$

Put the value into the formula

$F=4\times4.9\times10^{-3}\times9.8$

$F=0.192\ N$

Hence, (i), The mass of two washers is 0.0098 kg.

(ii). The applied force on the car is 0.192 N.

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3 years ago

A magnesium oxide component must not fail when a tensile stress of 10.5 MPa is applied. Determine the maximum allowable surface

Aloiza [94]

Answer:

Maximum permitted surface crack length is 1.29 mm

Explanation:

As per the question:

Tensile stress, $\sigma = 10.5\ MPa = 10.5\times 10^{6}\ Pa$

Surface energy of magnesium oxide, SE = $1.0\ J/m^{2}$

Modulus of elasticity of the material, E = 225 GPa = $225\times 10^{9}\ Pa$

Now,

To calculate the maximum allowable surface crack length:

$L = \frac{2E\times SE}{\sigma^{2}\pi }$

$L = \frac{2\times 225\times 10^{9}\times 1.0}{10.5\times 10^{6}\times \pi } = 1.29\times 10^{- 3}\ m = 1.29\ mm$

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3 years ago

an engineer is developing a conveyor belt to help move heavy boxes in factories. The device should be able to move boxes from th

Elena-2011 [213]

Answer:

This conveyor belt should be connected to fixed pulleys in case their total effort is less than the load. But if it's not that, the belt should be connected to movable pulleys.

Explanation:

$.$

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2 years ago

an oscillator with a mass of 310 g and a period of 0.180 s has amplited that decreases by 20 % during each complete oscillation

Alex

Answer:

1.241/s

Explanation:

From this question, we have the following information:

Mass of oscillator = 310

The time Period, t = 0.180

Decrease = 20 %

Amplitude of oscillation has the formula

A = Aoe^yt

1-20% = 80%

= 0.80

Y x 0.18 = -ln(0.8)

= 0.18Y = 0.2234

Y = 0.2233/0.18

Y = 1.241/s

The damping constant y is Therefore equal to 1.241s

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3 years ago

A 16 kg science book is dropped of a 120 meter high cliff. Assuming a closed system:

Lapatulllka [165]

Answer:

Explanation:

The mass of that science book...wow. In pounds that would be 35.2! Yikes!

Anyway, we need final velocity here, and the mass of the book has nothing to do with how fast it falls. Everything is pulled by the same gravity. A feather falls at 9.8 m/s/s and so does an elephant. Mass is useless information. The equation we will use is

$v^2=v_0^2+2a$ Δx where

v is the final velocity, our unknown,

v₀ is the initial velocity which is 0 since someone had to be holding the book before dropping it,

a is the pull of gravity which is always -9.8 m/s/s, and

Δx = -120 which is the displacement (it's negative because the book falls below the point from which it was dropped). Filling in:

$v^2=0^2+2(-9.8)(-120)$ so

$v=\sqrt{2(-9.8)(-120)}$ and

v = 48 m/s

As far as how far above the bottom of the cliff the object is when it is moving at 12 m/s we will use the same equation, but the velocity will be 12:

$12^2=0^2+2(-9.8)$ Δx and

144 = -19.6Δx so

Δx = -7.3 m. That's how far from the top of the cliff it is. We subtract then t find out how far it is from the bottom:

120 - 7.3 = 112.7 m off the ground.

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3 years ago