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VladimirAG [237]
3 years ago
15

suppose that the earth were moved to a distance of 0.25 AU from the sun. How much stronger or weaker would the suns gravitationa

l pull on earth?explain​
Physics
1 answer:
Leviafan [203]3 years ago
3 0

Answer:

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Explanation:

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You might be interested in
A mass on a horizontal surface is connected to the spring and pulled to the right along the surface stretching the spring by 25
solniwko [45]

Answer:

320 N/m

Explanation:

From Hooke's law, we deduce that

F=kx where F is applied force, k is spring constant and x is extension or compression of spring

Making k the subject of formula then

k=\frac {F}{x}

Conversion

1m equals to 100cm

Xm equals 25 cm

25/100=0.25 m

Substituting 80 N for F and 0.25m for x then

k=\frac {80}{0.25}=320N/m

Therefore, the spring constant is equal to 320 N/m

3 0
4 years ago
Draw vectors to represent these motions:<br>moving 3 cm to the right<br>moving 2 cm to the left
Ira Lisetskai [31]
Vector. Oh yeeeeeeeeahhh
7 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
A 40-kg worker climbs a ladder upwards for 15m. What work was done during their climb upwards?
Bingel [31]

Answer:

Explanation:

The work increased the potential energy

W = PE = mgh = 40(9.8)(15) = 5880 J(oules)

4 0
2 years ago
Which step should you take before you begin a scientific investigation?
Hoochie [10]

Answer:

The correct answer is C.

Clarify any confusing information in the instructions.

Explanation:

The very first thing that we need before conducting a scientific investigation is a set of instructions which might include the following information:

-What we are experimenting on?

-How are we going to do it?

-What things are needed for it?

If there are any confusions in the instructions, the chances of making a mistake rises. So, it is very important to clarify the confusions found in the instructions before choosing things needed for the experiment and starting the experiment.

6 0
3 years ago
Read 2 more answers
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