ᴛᴡᴏ ᴄʜᴀᴍʙᴇʀꜱ ᴡɪᴛʜ ᴛʜᴇ ꜱᴀᴍᴇ ꜰʟᴜɪᴅ ᴀᴛ ᴛʜᴇɪʀ ʙᴀꜱᴇ ᴀʀᴇ ꜱᴇᴘᴀʀᴀᴛᴇᴅ ʙʏ ᴀ 30-ᴄᴍ-ᴅɪᴀᴍᴇᴛᴇʀ ᴘɪꜱᴛᴏɴ ᴡʜᴏꜱᴇ ᴡᴇɪɢʜᴛ ɪꜱ 25 ɴ, ᴀꜱ ꜱʜᴏᴡɴ ɪɴ ᴛʜᴇ ꜰɪɢᴜʀᴇ. ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴛʜᴇ ɢᴀɢᴇ ᴘʀᴇꜱꜱᴜʀᴇ (ɪɴ ᴋᴘᴀ) ɪɴ ᴄʜᴀᴍʙᴇʀ ᴀ.

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Provide a complete solution and answer for the given problem in the attached image.

please show the work in handwritten format only ,Kindly do not type. The answer is already provided

a) A locomotive is travelling at 35 m/s up an incline of 0.006 radians when the engine cuts out. How far will the locomotive travel up the incline before coming to rest? Assume there are no frictional losses, no energy input from the engine, and take acceleration due to gravity to be 9.81 m/s². Give your answer in km. Round your answer to 4 significant figures. b) What is the speed (in m/s) of the locomotive after it has travelled a distance 5.83 km? Round your answer to 3 significant figures.

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a) Fig. 2 shows a rim of a tractor tyre of diameter 60 cm and width 25 cm. After inspection, the tyre was found imbalanced. It seems to the equivalent a 2 kg mass 5 cm from the outside of the tyre and 30cm from the centre of the tyre at 100°, and another one 1 kg mass at 18 cm from outside of the tyre and 10 cm from centre of the tyre at 300° angle. Find masses that should be added to make it balanced at the outside and inside tyre at the rim (30cm from the centre) B 5cm 2kg r3-30cm 18 cm Outside YU с Fig. 2 10cm | by Inside 18cm 300° 100⁰ 10 cm с

15 please need help, keep in mind the 7 basic units and watch the sig figs thank you so much.

Fora.10 H/p6 fit:Complete the following table. Hole Shaft Tolerance: Tolerance: Limit deviations Upper deviation (ES): Lower deviation (EI): Upper deviation (ES): Lower deviation (El): Limit of size Maximumlimit of size: Minimum limit of size: Maximum limit of size: Minimum limit of size: FIT Type of Fit: Мах Mini. Go-No GO Gages for the above Hole Complete the following table. Show your calculation and draw Go -Not Go representation. Go Plug Gage NOT Go Plug Gage Nominal size: Nominal size: Tolerance: Tolerance: Class: Max limit of size: Class: Max limit of size: Min limit of size: Min limit of size:

The left side of this equation tells how much energy Q the cylinder gives to the water while it cools. The right side of this equation tells how much energy Q the water and aluminum cup absorb from the cylinder to warm up. Because it is the same energy, they are equal. What is known in this equation? Mcyl 411.7 g, malum 46.5 g, malum+water = 175 g Can you find: mwater =? g Twater = Talum = 20°C (water and cup of room temperature) 90°C, T; = 35°C (hot cylinder and cool "cylinder+cup+water" temperatures) Tcyl kCal Calum = 0.22, Cwater 1 (specific heat of water and aluminum, measured in units kg-°C What are we looking for is Ccul - How we find it? Plug all the numbers into the equation (1), Ccul will be one unknown which you can calculate from the equation. Important, convert all the masses from grams to kilograms! After you find Ccyl, compare it to known value for the copper 0.093(our cylinder is made out of copper). |Ceyl -0.093| % : · 100% 0.093