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3 years ago

If an astronaut throws an object in space, the object’s speed will _____

1 answer:

7 0

The object's speed will not change.

In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
$\sum F=ma$
where the first term is the resultant of the forces acting on the body, m is the mass of the object and a its acceleration, we see that if no forces act on the object, then the acceleration is zero. Therefore, the acceleration of the object is zero, and its velocity remains constant.

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Determine the voltage ratings of the high-and-low voltage windings for this connection and the MVA rating of the autotransformer

SIZIF [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The High Voltage Rating for Auto - Transformer is 86kV

The Low Voltage Rating for Auto - Transformer is 78kV

The MVA rating is 268.75 $MVA$

The efficiency is 99.4%

Explanation:

From the question we are given are given that

The transformer has Mega Volt Amp rating of 25MVA

The frequency is 60-Hz

Voltage rating 8.0kV : 78kV

The short circuit test gives : 453kV,321A,77.5kW

The open circuit test gives : 8.0kV, 39.6A, 86.2kW

This can be represented on a diagram shown on the second uploaded image

From this diagram we can deduce that the The High Voltage Rating for Auto - Transformer is 86kV and the Low Voltage Rating for Auto - Transformer is 78kV

Now to obtain the current flowing through the 8kV coil in the Auto-transformer we have

$\frac{25 \ Mega \ Volt\ Ampere }{8\ Kilo Volt}$

The volt will cancel each other

$\frac{25*10^6}{8*10^3} = 3125\ A$

Now to obtain the required MVA rating we would multiply the value of Power obtained during the open circuit test by the value of the current calculated.we are making use of the power obtain during open circuit testing because the transformer at this point is not under any load.

$MVA \ rating = (86*10^3)(3125) =268.75$

We need to understand that Iron losses is due to open circuit test which has power = 86.2kW

While copper loss is due to short circuit test which has power = 77.5kW

The the current flowing through the secondary coil $I_2$ as shown in the circuit diagram can be obtained as

$I_2 = \frac{25*10^6}{78*10^3} =320.52 A \approx 321$

Now the efficiency can be obtained as thus

$\frac{(operational \ MVA )*(Power factor \pf))}{(operational\ MVA (power factor pf) + copper loss + Iron loss)}*\frac{100}{1}$

=99.941%

8 0

2 years ago

Any one their to help

Dmitry [639]

Answer:

five dollars

Explanation:

im thick lol

7 0

2 years ago

On a horizontal surface is located

Ierofanga [76]

By Newton's second law, the net vertical force acting on the object is 0, so that

n - w = 0

where n = magnitude of the normal force of the surface pushing up on the object, and w = weight of the object. Hence n = w = mg = 196 N, where m = 20 kg and g = 9.80 m/s².

The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of static friction µ is such that

80 N = µ (196 N) → µ = (80 N)/(196 N) ≈ 0.408

Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of kinetic friction ν is

40 N = ν (196 N) → ν = (40 N)/(196 N) ≈ 0.204

And so the closest answer is C.

(Note: µ and ν are the Greek letters mu and nu)

3 0

3 years ago

Narysuj wykres zależności v(t) jeśli w chwili początkowej t=0 V=10m/s w każdej sekundzie szybkość zmniejsza się o 1m/s . Po jaki

irina1246 [14]

1) See graph in attachment

2) 10 s

3) 50 m

Explanation:

In this problem, we have an object initially moving with a velocity of

v = 10 m/s

when the time is

t = 0 s

Then, we are told that the speed of the object is decreasing by 1 m/s every second. This means that on a velocity-time graph, the motion will be represented by a straight line, starting from v = 10 when t = 0, and decreasing by 1 m/s every second.

The result can be found in the graph in attachment.

Moreover, we can also infer that the motion of the object is accelerated (because velocity is changing), and that the acceleration is constant and it is equal to

$a=1 m/s^2$