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3 years ago

10

A machine produces a larger force than you exert to operate the machine. How does this input distance of the machine compare to

its output distance?

1 answer:

arlik [135]3 years ago

7 0

Answer:

Because the output force is greater than the input force, the input distance must be greater than the output distance.

Explanation:

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3 years ago

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Susan and Hannah are each riding a swing. Susan has a mass of 25 kilograms, and Hannah has a mass of 30 kilograms. Susan’s swing

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Speed has a greater effect than mass

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Water (density = 1 ´ 103 kg/m3) flows at 10 m/s through a pipe with radius 0.030 m. the pipe goes up to the second floor of the

Hatshy [7]

density of water = $1000 kg/m^3$

velocity of flow = $10 m/s$

radius of pipe = $0.030 m$

Height of second floor = $2 m$

Now we can use here Bernuoli's Equation to find the speed of water flow at second floor

$P_1 + 1/2\rho v_1^2 + \rho g h_1= P_2 + 1/2 \rho v_2^2 + \rho g h_2$

$P + 1/2 * 1000 * 10^2 + 1000* 9.8 * 0 = P + 1/2 * 1000 * v^2 + 1000*9.8*2$

$v = 7.8 m/s$

Now in order to find the radius of pipe we can use equation of continuity

$A_1 v_1 = A_2 v_2$

$\pi *0.030^2 * 10 = \pi * r^2 * 7.8$

$r = 0.034 m$

So radius of pipe at second floor is 0.034 meter

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3 years ago

Which of the following is an example of an endothermic reaction?

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Answer:

D, photosynthesis

Explanation:

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2 years ago

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2. One mole of a monatomic ideal gas undergoes a reversible expansion at constant pressure, during which the entropy of the gas

Anna35 [415]

Answer:

The initial and final temperatures of the gas is 300 K and 600 K.

Explanation:

Given that,

Entropy of the gas = 14.41 J/K

Absorb gas = 6236 J

We know that,

$ds=\dfrac{dQ}{dt}$

At constant pressure,

$dQ=C_{p}dt$

$\Delta s=\int_{T_{1}}^{T_{2}}{\dfrac{C_{p}dT}{T}}$

$\Delta s=C_{p}ln\dfrac{T_{2}}{T_{1}}$

Put the value into the formula

$14.41=2.5\times8.3144(ln\dfrac{T_{2}}{T_{1}})$

$\dfrac{14.41}{2.5\times8.3144}=ln\dfrac{T_{2}}{T_{1}}$

$0.693=ln\dfrac{T_{2}}{T_{1}}$

$ln2=ln\dfrac{T_{2}}{T_{1}}$

$T_{2}=2T_{1}$ ...(I)

We need to calculate the initial and final temperatures of the gas

Using formula of energy

$\Delta Q=C_{p}\Delta T$

Put the value into the formula

$6236=2.5\times8.3144(T_{2}-T_{1})$

$6236=20.786(T_{2}-T_{1})$

$T_{2}-T_{1}=\dfrac{6236}{20.786}$

$T_{2}-T_{1}=300$

Put the value of T₂

$2T_{1}-T_{1}=300$

$T_{1}=300\ K$

Put the value of T₁ in equation (I)

$T_{2}=2\times300$

$T_{2}=600\ K$

Hence, The initial and final temperatures of the gas is 300 K and 600 K.

8 0

3 years ago