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3 years ago

How many sides and points do a hexagon have

Physics

2 answers:

tamaranim1 [39]3 years ago

8 0

A hexagon has 6 sides

MAVERICK [17]3 years ago

8 0

Hexagon has 6 points

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Which statements describe the sun? Check all that apply

lesya692 [45]

Answer:

All but 4 I believe

Explanation:

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3 years ago

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Brainliest if correct

ozzi

Answer:

It's is c coz the weight is equal to the normal force

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3 years ago

The 6th harmonic of a pipe is open at both ends has a frequency of 732 Hz, on a day when the temperature is 10 degrees Celsius.

Stels [109]

Answer:

안녕하세요 우리는 어떻게 응을 하루?

Explanation:

이봐 우리는 오늘 응

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3 years ago

Oxygen at 50 lbf/in.2, 200 F is in a piston/cylinder arrangement with a volume of 4 ft3. It is now compressed in a polytropic pr

Ksju [112]

Answer:

The value of heat transfer during the process Q = - 29.49 KJ

Explanation:

Given data

$P_{1}$ = 50 $\frac{lbf}{in^{2} }$ = 344.7 k pa

$V_{1} = 4 ft^{3}$ = 0.113 $m^{3}$

$T_{1} = 200$ F = 366.4 K

$T_{2} = 400 F$ = 477.6 K

Poly tropic index n = 1.2

gas constant for oxygen = 0.26 $\frac{KJ}{kg K}$

From ideal gas equation

$P_{1}$ $V_{1}$ = m R $T_{1}$

Put all the values in above equation we get

⇒ 344.7 × 0.113 = m × 0.26 × 366.4

⇒ m = 0.408 kg

Heat transfer in poly tropic process is given by

Q = $\frac{\gamma - n}{( \gamma - 1)( n - 1)} [ {m R (T_{1} - T_{2} ) ]$

Put all the values in above formula we get

⇒ Q = $\frac{1.4 - 1.2}{( 1.4 - 1)( 1.2 - 1)} [ {m R (T_{1} - T_{2} ) ]$

⇒ Q = 2.5 × 0.408 × 0.26 × ( 366.4 - 477.6 )

⇒ Q = - 29.49 KJ

This is the value of heat transfer during the process & negative sign shows that heat is lost during the process.

3 0

4 years ago

Consider two laboratory carts of different masses but identical kinetic energies and the three following statements. I. The one

kolbaska11 [484]

Answer:

d) I and III only.

Explanation:

Let be $m_{1}$ and $m_{2}$ the masses of the two laboratory carts and let suppose that $m_{1} > m_{2}$ . The expressions for each kinetic energy are, respectively:

$K = \frac{1}{2}\cdot m_{1}\cdot v_{1}^{2}$ and $K = \frac{1}{2}\cdot m_{2}\cdot v_{2}^{2}$ .

After some algebraic manipulation, the following relation is constructed:

$\frac{m_{1}}{m_{2}} = \left(\frac{v_{2}}{v_{1}}\right)^{2}$

Since $\frac{m_{1}}{m_{2}} > 1$ , then $\frac{v_{2}}{v_{1}} > 1$ . That is to say, $v_{1} < v_{2}$ .

The expressions for each linear momentum are, respectively:

$p_{1} = \frac{2\cdot K}{v_{1}} = m_{1}\cdot v_{1}$ and $p_{2} = \frac{2\cdot K}{v_{2}} = m_{2}\cdot v_{2}$

Since $v_{1} < v_{2}$ , then $p_{1} > p_{2}$ . Which proves that statement I is true.

According to the Impulse Theorem, the impulse needed by cart I is greater than impulse needed by cart II, which proves that statement II is false.

According to the Work-Energy Theorem, both carts need the same amount of work to stop them. Which proves that statement III is true.

6 0

3 years ago