Find the missing number for each unit rate? 10/2 =?/1 and 16/4=?/1

A result of chemical change is

Tcecarenko [31]

We're u can never put it back together

4 0

3 years ago

Calculate the moment of 3kg and 20cm away from the pivot

Oksi-84 [34.3K]

Answer:

5.88 Nm

Explanation:

Applying,

Moment(M) = Force(F)×perpendicular distance(d)

M = F×d.................... Equation 1

But,

F = mg.................. Equation 2

Where m = mass, g = acceleration due to gravity.

Substitute equation 2 into equation 1

M = mgd................. Equation 3

From the question,

Given: m = 3 kg, d = 20 cm = 0.2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 3

M = 3×0.2×9.8

M = 5.88 Nm

4 0

3 years ago

A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160

Marianna [84]

Answer:

(a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

Explanation:

Given that,

Sample of an ideal gas = 0.500 mol

Initial pressure = 400 kPa

Final pressure = 160 kPa

Temperature = 300 K

(a) for isothermal,

Temperature will be same.

We need to calculate the volume of gas

$P_{f}V_{f}=P_{i}V_{i}$

$V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}$

Put the value into the fomrula

$V_{f}=(\dfrac{400}{160})V_{i}$

$V_{f}=2.5 V_{i}$

We need to calculate the work done

Using equation of energy

$dQ=dW$

$dQ=nRTln(\dfrac{P_{in}}{P_{f}})$

$dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})$

$dQ=1143\ J$

(b). For adiabatic,

No transfer of heat between system and surroundings

We need to calculate the final temperature

Using formula of gas

$P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}$

$T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}$

Put the value into the formula

$T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300$

$T_{f}=207.9\ K$

We need to calculate the wok done in adiabatic

Using formula of work done

$W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}$

$W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}$

$W=574\ J$

Hence, (a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

6 0

3 years ago

What happens when one force acts on an object?

Pachacha [2.7K]

Answer:

if the weight of the object is equal to the force applied, there will be no change

if the force applied is less than the weight of the object, there won't be any change

if the force application ed is greater than the weight of the object, than the object will move.

Explanation:

here the weight referred to the force of the object.

5 0

3 years ago

Convert 500ml to how many liters

Mashutka [201]

500 ml = 0.5 liters. that's what i'm getting

hope it helps

7 0

3 years ago

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