<h3>
Answer:</h3>
5.5 N
<h3>
Explanation:</h3>
<u>We are given;</u>
- Real weight of an object in air as 10 N
- Apparent weight in water as 7 N
- Relative density of a liquid is 1.5
We need to calculate the apparent weight when in liquid .
- First we calculate upthrust
Upthrust = Real weight - Apparent weight
= 10 N - 7 N
= 3 N
- Then calculate the upthrust in the liquid.
we need to know that;
Relative density of a liquid = Upthrust in liquid/Upthrust in water
Therefore;
1.5 = U ÷ 3 N
Upthrust in liquid = 3 N × 1.5
= 4.5 N
- Therefore, the upthrust of the object in the liquid is 4.5 N
But, Upthrust = Real weight - Apparent weight
Therefore;
Apparent weight in Liquid = Real weight - Upthrust
= 10 N - 4.5 N
= 5.5 N
Thus, the weight when the object is immersed in the liquid is 5.5 N
Answer:
526.57 Pa
Explanation:
P ( pressure at the bottom of the container) = 1.049 × 10^5 pa
Using the formula of pressure in an open liquid
Pw ( pressure due to water) = ρhg where ρ is density of water in kg/m³, h is the height in meters, and g is acceleration due to gravity in m/s²
Pw = 1000 × 9.81 ×0.209 = 2050.29 Pa
P( atmospheric pressure) = 1.013 × 10^5 Pa
Pl ( pressure due to the liquid) = ρ(density of the liquid) × h (depth of the liquid) × g
Subtract each of the pressure from the absolute pressure at the bottom
P(bottom) - atmospheric pressure
(1.049 × 10^5) - (1.013 × 10^5) = 0.036 × 10^5 = 3600 Pa
subtract pressure due to water from the remainder
3600 - 2050.29 = 1549.71 Pa
1549.71 = ρ(density of the liquid) × h (depth of the liquid) × g
ρ (density of the liquid) = 1549.71 / (h × g) = 1549.71 / (0.3 × 9.81) =526.57 Pa
Answer:
Meters
Explanation:"How FAR did the athlete run?"
Also it talked about meters
