A person holds a portable fire extinguisher that ejects 1.0kg of water per second horizontally at a speed of 6.0 m/s. What newto
1 answer:
You might be interested in
Answer:
The frequency of the oscillation is 2.45 Hz.
Explanation:
Given;
mass of the spring, m = 0.5 kg
total mechanical energy of the spring, E = 12 J
Determine the spring constant, k as follows;
E = ¹/₂kA²
kA² = 2E
k = (2E) / (A²)
k = (2 x 12) / (0.45²)
k = 118.519 N/m
Determine the angular frequency, ω;
Determine the frequency of the oscillation;
ω = 2πf
f = (ω) / (2π)
f = (15.396) / (2π)
f = 2.45 Hz
Therefore, the frequency of the oscillation is 2.45 Hz.
Answer:
a) 3.37 x
b) 6.42kg/
Explanation:
a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .
Weight of metal in air = 50N = mg implies the mass of metal is 5kg.
Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x
. So density of metal = mass of metal / volume of metal = 5 / 14 x
= 3.37 x
b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/
Answer:
the final pressure of the gas is 60 kPa.
Explanation:
Given;
initial pressure of the gas, P₁ = 50 kPa = 50,000 Pa
initial temperature of the gas, T₁ = 27⁰ C = 27 + 273 = 300 k
final temperature of the gas, T₂ = 87⁰ C = 87 + 273 = 360 K
Let the final pressure of the gas = P₂
Apply pressure law;
Therefore, the final pressure of the gas is 60 kPa.