The mass (in kg) of water in the pool is 1.264 kg.
<h3>What is mass?</h3>
A numerical measurement of inertia, a basic characteristic of all matter, in physics. It essentially refers to a body of matter's resistance to changing its speed or location in response to the application of a force. The change caused by an applied force is smaller the more mass a body has.
Mass is a physical body's total amount of matter. Inertia, or the body's resistance to acceleration when a net force is applied, is also measured by this term. The strength of an object's gravitational pull to other bodies is also influenced by its mass. The kilogram serves as the SI's fundamental mass unit.
Given: A pool is 31.4 m long and 40.0 m wide. if the average depth of water is 3.30 ft
31.4 40.0 = 1256
1 m = 39.37 in = 3.28 ft
3.30 ft = 1.01 m
Volume of pool = 1256 1.01 = 1264
Since 1 water contains the volume of the pool is
1.264 E3 E3 = 1.264E6 kg
(or 1.264 kg)
The mass (in kg) of water in the pool is 1.264 kg.
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Answer:
877
Explanation:
P1 = 150 atm
V1 = 0.12 m^3
P2 = 1.2 atm
V2 = ?
radius of each balloon, r = 0.16 m
Volume of each balloon,
V2 = 0.0171 m^3
Let the number of balloons be N.
So,
P1 x V1 = N x P2 x V2
150 x 0.12 = N x 1.2 x 0.0171
N = 877.19
So, the number of balloons be 877.
<span>combustion will not burn</span>
Answer: Its A because i had the same question and it was a
The crate is in equilibrium. Newton's second law gives
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0
∑ <em>F</em> (horizontal) = <em>p</em> - <em>f</em> = 0
where
• <em>n</em> = magnitude of the normal force
• <em>mg</em> = weight of the crate
• <em>p</em> = mag. of push exerted by movers
• <em>f</em> = mag. of kinetic friciton, with <em>f</em> = 0.60<em>n</em>
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It follows that
<em>p</em> = <em>f</em> = 0.60<em>mg</em> = 0.60 (43.0 kg) <em>g</em> = 252.84 N
so that the movers perform
<em>W</em> = <em>p</em> (10.4 m) ≈ 2600 J
of work on the crate. (The <em>total</em> work done on the crate, on the other hand, is zero because the net force on the crate is zero.)