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alex41 [277]
3 years ago
14

Find the impulse of a 50. kg object under the following scenarios:

Physics
1 answer:
Alisiya [41]3 years ago
7 0
B. The object stops from a velocity of 12.0 m/s
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Loud of dust and gas in space is a(n) Blank Space __________.
madreJ [45]
A cloud in space that is composed of dust and gas is call a nebula. The word "nebula" is derived from the Latin word for "cloud," and nebulae in space are indeed large interstellar clouds made up of dust, hydrogen, helium and plasma.

6 0
3 years ago
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Which law of physics relates electric fields and current
riadik2000 [5.3K]

Answer:

Ohms law

Explanation:

Which states that the current flowing through any cross-section of the conductor is directly proportional to the potential differenceapplied across its end, provided physical conditions like temperature and pressure remain constant.

7 0
4 years ago
What are the equivalent celsius and kelvin temperatures of -128 f?
Gre4nikov [31]

Answer:

128 Kelvin = 128 - 273.15 = -145.15 Celsius. Temperature conversion chart Sample temperature conversions 103.55 Kelvin to degrees Fahrenheit 39.82 degrees Fahrenheit to Kelvin

Explanation:

hope this helps have a good day

6 0
2 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
4 years ago
The density of an object is dependent upon the object’s mass and ---
kotegsom [21]

Answer:Volume

Explanation:

Density = mass/ Volume

7 0
3 years ago
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