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kumpel [21]
3 years ago
13

Water enters a 2 m3 tank at a rate of 6 kg/s and is withdrawn at a rate of 2 kg/s. The tank is initially half full. What type of

process is this?
Physics
1 answer:
NemiM [27]3 years ago
8 0

Answer:

This process is semicontinuous.

Explanation:

Given that,

Volume = 2 m³

Enter flow rate = 6 kg/s

Exit flow rate = 2 kg/s

The tank is initially half full.

We need to find what type of process

Using given data,

This process is not continuous because given enter and exit flow rate is not equal.

This process is semicontinuous and the water level in the tank does not reach a constant level.

Hence, This process is semicontinuous.

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In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
3 years ago
If an electronin an electron beam experiences a downward force of 2.0x10^-14N while traveling in a magnetic field of 8.3x10^-2T
Anni [7]

Answer:

Explanation:

Given that,

Force is downward I.e negative y-axis

F = -2 × 10^-14 •j N

Magnetic field is westward, +x direction

B = 8.3 × 10^-2 •i T

Charge of an electron

q = 1.6 × 10^-19C

Velocity and it direction?

Force in a magnetic field is given as

F = q(V×B)

Angle between V and B is 270, check attachment

The cross product of velocity and magnetic field

F =qVB•Sin270

2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2

Then,

v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)

v = 1.51 × 10^6 m/s

Direction of the force

Let x be the direction of v

-F•j = v•x × B•i

From cross product

We know that

i×j = k, j×i = -k

j×k =i, k×j = -i

k×i = j, i×k = -j OR -k×i = -j

Comparing -k×i = -j to given problem

We notice that

-F•j = q ( -V•k × B×i)

So, the direction of V is negative z- direction

V = -1.51 × 10^6 •k m/s

6 0
3 years ago
All of the following are locations for kinesthetic system receptors except __________. A. muscles B. semicircular canals C. join
jasenka [17]

Answer:

I think it's B

Explanation:

Because hese receptors are found in muscles, tendons, and joints.

5 0
3 years ago
Read 2 more answers
ਭਾਰਤ ਵਿੱਚ ਸਭ ਤੋਂ ਵੱਧ ਵਰਖਾ ਕਿੱਥੇ<br>ਹੁੰਦੀ ਹੈ।​
astra-53 [7]

Answer:

I don’t understand:(

3 0
3 years ago
) you carry a 7.0 kg bag of groceries 1.2 m above the ground at constant velocity across a 2.7 m room. how much work do you do o
lapo4ka [179]
M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries

The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.

At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
    = (7.0 kg)*(9.8 m/s²)*(1.2 m)
    = 82.32 J

The change in PE is equal to the work done.

Answer: 82.3 J

3 0
3 years ago
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