Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located
1 answer:
Answer:43.33 cm mark
Explanation:
Given
mass 1 is located at the 10 cm mark with weight of 15 kg
mass 2 is located at 60 cm mark with weight of 30 kg
string should be attached between them to balance the system
so the distance between the the two masses is 50 cm
For system to be balance torque of both the weight must nullify each other
Let us suppose string is at a distance of x cm from 15 kg mass so 30 kg mass is at a distance of 50-x cm
Balancing torque
so string should be at a mark of 10+33.33=43.33 cm
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Answer:
We need to multiply 12 to each term to eliminate fractions.
Explanation:
Given expression:
To eliminate the fraction we need to multiply each term by least common multiple of the denominators of the fraction.
The denominators in the above expressions are:
4, 3 and 2
The multiples of each can be listed below.
2⇒ 2,4,6,8,10,<u>12</u>,14,16.....
3⇒ 3,6,9,<u>12</u>,15,18
4⇒ 4,8,<u>12</u>.......
From the list of the multiples stated, we can see the least common multiple is 12.
So we will multiply each term by 12.
Multiplying 12 to both sides.
Using distribution,
Thus we successfully eliminated the fractions.
Explanation:
The given data is as follows.
mass, m = 75 g
Specific heat of water = 4.18
First, we will calculate the heat required for water is as follows.
q =
=
= 8464.5 J/mol
= 8.46 kJ ......... (1)
Also, it is given that = (20 + 273) K = 293 K and specific heat of ice is 2.108 kJ/kg K.
Now, we will calculate the heat of fusion as follows.
q =
=
= -46.32 kJ ......... (2)
Now, adding both equations (1) and (2) as follows.
8.46 kJ - 46.32 kJ
= -37.86 kJ
Therefore, we can conclude that energy in the form of heat (in kJ) required to change 75.0 g of liquid water at to ice at
is -37.86 kJ.