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2 years ago

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The specific heat capacity of oxygen is 918 J/kg °C. That means that it takes 918 J of energy to raise _____ kg of oxygen by 1°C

. What number completes the sentence?

1 answer:

Anarel [89]2 years ago

4 0

Answer:

Explanation:

1 kg

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A 30-km, 34.5-kV, 60-Hz, three-phase line has a positive-sequence series impedance z 5 0.19 1 j0.34 V/km. The load at the receiv

zmey [24]

Answer:

(a) With a short line, the A,B,C,D parameters are:

A = 1pu B = 1.685∠60.8°Ω C = 0 S D = 1 pu

(b) The sending-end voltage for 0.9 lagging power factor is 35.96 $KV_{LL}$

(c) The sending-end voltage for 0.9 leading power factor is 33.40 $KV_{LL}$

Explanation:

(a)

Considering the short transition line diagram.

Apply kirchoff's voltage law to the short transmission line.

Write the equation showing the relations between the sending end and the receiving end quantities.

Compare the line equations with the A,B,C,D parameter equations.

(b)

Determine the receiving-end current for 0.9 lagging power factor.

Determine the line-to-neutral receiving end voltage.

Determine the sending end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

(c)

Determine the receiving-end current for 0.9 leading power factor.

Determine the sending-end voltage of the short transition line.

Determine the line-to-line sending end voltage which is the sending end voltage.

8 0

2 years ago

A space vehicle deploys its re–entry parachute when it's traveling at a vertical velocity of –150 meters/second (negative becaus

dexar [7]

Answer:

$a=5m/s^2$

Explanation:

Aceleration is a change on the velocity of the object in a given time.

For this case: the initial velocity is

$v_{1}=150m/s$

and the final velocity is :

$v_{2}=0 m/s$

so, the change in velocity is:

$\Delta v =v_{2}-v_{1}=0m/s - (-150m/s) = 150 m/s$

and the change in time , according to the problem:

$\Delta t=30s$

So, the aceleration is:

$a=\frac{\Delta v}{\Delta t} = \frac{150m/s}{30s} = 5m/s^2$

6 0

2 years ago

A man can row at 6kmhr in still water and want to cross a river to a position exactly opposite his starting point. If the river

Lina20 [59]

Answer:

if the river is 5km wide and is flowing at 4km/hr. eastwards find by scale ... Find either by scale drawing or by calculation (1) the direction in which he must ... He could row his boat directly across the river to point C and then run to B, or he ... A man who can swim at 5km/h in still water swims towards the east to cross arriver.

Explanation:

3 0

2 years ago

A small light cylinder and a large heavy cylinder are released at the same time and roll down a ramp without slipping. Which one

forsale [732]

Answer:

C. Both reach the bottom at the same time.

Explanation:

For a rolling object down an inclined plane , the acceleration is given below

a = g sinθ / (1 + k² / r² )

θ is angle of inclination , k is radius of gyration , r is radius of the cylinder

For cylindrical object

k² / r² = 1/2

acceleration = g sinθ /( 1 + 1/2 )

= 2 g sinθ / 3

Since it does not depend upon either mass or radius , acceleration of both the cylinder will be equal . Hence they will reach the bottom simultaneously.

6 0

2 years ago

A ski lift is used to transport people from the base of a hill to the top. If the lift leaves the base at a velocity of 15.5 m/s

Neko [114]

This can be calculated with the law of conservation of energy. The sky lift is starting with the speed v= 15.5 m/s and all of it's kinetic energy Ek is transformed to potential energy Ep so the energies have to be equal: Ep=Ek.
Since Ek=(1/2)*m*v² where m is mass and v is the speed, Ep=m*g*h, where m is mass, g= 9.81 m/s² and h is height. Now:

Ek=Ep

(1/2)*m*v²=m*g*h, masses cancel out,

(1/2)*v²=g*h, divide by g to get the height,

(1/2*g)*v²=h and now plug in the numbers:

h=12.245 m. Height of the hill rounded to the nearest tenth is h=12.25 m

5 0

2 years ago

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