The specific gravity of the object’s material is 5.09.
<h3>To calculate the specific gravity of the object:</h3>
Weight difference = 9 - 7.2 = 1.8 N = Buoyant force of water
Buoyant Force in water(Fb) = density of water x g x volume of the body(Vb)
1.8 = 1000 x 9.81 x Vb
Vb = 1.8/9810 cubic meter
Now, in the air;
Weight of body = mg = 9 N
Mass of body,m = 9/9.81 Kg
So,
Density of body = m/ Vb
= 9/9.81 ÷ 1.8/9810
= 5094.44 kg per cubic meter
The specific gravity of body = density of body ÷ density of water
= 5094.44 ÷ 1000
= 5.09
Therefore, Specific gravity of body = 5.09
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Answer:
I'd say 85km sorry if wrong
Explanation:
6 m/s because the position of the object was increasing 6 m every second
Answer: 2.37N
Explanation:
According to coulombs law which states that the force of attraction (F) between two charges (q1 and q2) is directly proportional to the product of their charges and inversely proportional to the square of the distance (r) between them. Mathematically,
F = kq1q2/r²
For the first two charges that are sitting 1.5 m apart with a force of 3 N between them, we have
3 = kq1q2/1.5²
3 = kq1q2/2.25
Kq1q2= 6.75... (1)
If the charges are now moved farther apart 2.25 m and one of the charges is increased by a factor of 4. The formula becomes
F2 = k(4q1)q2/2.25² (q1 has been increased by factor of 4)
k(4q1)q2 = 5.06F2 ... (2)
Dividing 2 by 1 we have
k(4q1)q2/kq1q2 = 5.06F2/3
4 = 5.06F2/3
5.06F2 = 12
F2= 12/5.06
F2 = 2.37N
Therefore the magnitude of the new force between the two charges is 2.37N
