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3 years ago

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Engineering

1 answer:

san4es73 [151]3 years ago

7 0

Answer:

I dont know. I have the same question. Can you help me out?

Explanation:

idk

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A weight-lifting athlete raises a mass of 160 kg through a vertical distance of 1.4 m. What force did

Over [174]

Answer:

1568N

2195.2J

Explanation:

Given parameters:

Mass of the weight = 160kg

Distance = 1.4m

Unknown:

Force applied to lift the weight = ?

Energy expended = ?

Solution:

The force applied in moving a body with a given mass through a distance is the weight;

Force applied = mg

Where m is the mass

g is the acceleration due to gravity

i. Applied force = 160 x 9.8 = 1568N

ii. The energy used to lift the weight is given as;

Energy = mgh

h is the vertical distance

Energy = 1568 x 1.4 = 2195.2J

8 0

3 years ago

A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude

Fofino [41]

Answer:

$u = 260.22m/s$

$S_{max} = 1141.07ft$

Explanation:

Given

$S_0 = 89.6ft$ --- Initial altitude

$S_{16.5} = 0ft$ -- Altitude after 16.5 seconds

$a = -g = -32.2ft/s^2$ --- Acceleration (It is negative because it is an upward movement i.e. against gravity)

Solving (a): Final Speed of the rocket

To do this, we make use of:

$S = ut + \frac{1}{2}at^2$

The final altitude after 16.5 seconds is represented as:

$S_{16.5} = S_0 + ut + \frac{1}{2}at^2$

Substitute the following values:

$S_0 = 89.6ft$ $S_{16.5} = 0ft$ $a = -g = -32.2ft/s^2$ and $t = 16.5$

So, we have:

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 32.2 * 16.5^2$

$0 = 89.6 + u * 16.5 - \frac{1}{2} * 8766.45$

$0 = 89.6 + 16.5u- 4383.225$

Collect Like Terms

$16.5u = -89.6 +4383.225$

$16.5u = 4293.625$

Make u the subject

$u = \frac{4293.625}{16.5}$

$u = 260.21969697$

$u = 260.22m/s$

Solving (b): The maximum height attained

First, we calculate the time taken to attain the maximum height.

Using:

$v=u + at$

At the maximum height:

$v =0$ --- The final velocity

$u = 260.22m/s$

$a = -g = -32.2ft/s^2$

So, we have:

$0 = 260.22 - 32.2t$

Collect Like Terms

$32.2t = 260.22$

Make t the subject

$t = \frac{260.22}{ 32.2}$

$t = 8.08s$

The maximum height is then calculated as:

$S_{max} = S_0 + ut + \frac{1}{2}at^2$

This gives:

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 32.2 * 8.08^2$

$S_{max} = 89.6 + 260.22 * 8.08 - \frac{1}{2} * 2102.22$

$S_{max} = 89.6 + 260.22 * 8.08 - 1051.11$

$S_{max} = 1141.0676$

$S_{max} = 1141.07ft$

Hence, the maximum height is 1141.07ft

8 0

3 years ago

50 points what shape is mars

Artyom0805 [142]

A sphere.............

5 0

3 years ago

Read 2 more answers

The volume fraction of particles in a WC-particle Cu-matrix CERMET is 0.84 Calculate the minimum expected elastic modulus, in GP

Lynna [10]

Answer:

the minimum expected elastic modulus is 372.27 Gpa

Explanation:

First we put down the data in the given question;

Volume fraction $V_f$ = 0.84

Volume fraction of matrix material $V_m$ = 1 - 0.84 = 0.16

Elastic module of particle $E_f$ = 682 GPa

Elastic module of matrix material $E_m$ = 110 GPa

Now, the minimum expected elastic modulus will be;

$E_{CT$ = ( $E_f$ × $E_m$ ) / ( $E_f$ $V_m$ + $E_m$ $V_f$ )

so we substitute in our values

$E_{CT$ = (682 × 110 ) / ( [ 682 × 0.16 ] + [ 110 × 0.84] )

$E_{CT$ = ( 75,020 ) / ( 109.12 + 92.4 )

$E_{CT$ = 75,020 / 201.52

$E_{CT$ = 372.27 Gpa

Therefore, the minimum expected elastic modulus is 372.27 Gpa

7 0

3 years ago

B is the desired product, and X and Y are foul pollutants that are expensive to get rid of. The specific reaction rates are at 2

AURORKA [14]

Answer:

See attachment for detailed answer.

Explanation:

Download pdf

3 0

3 years ago