3. Which is an example of a safe laboratory procedure?
1 answer:
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Answer:
96046 Ns.
Explanation:
We shall represent velocity in vector form considering east direction as + ve x axis and north as + y direction.
40 km/h in the east
V₁ = 40 i
V₂ = 50j
momentum p₁ = mV₁
= 1500 X 40 i
= 60000 i
Momentum p₂ = mV₂
= 1500 X 50j
= 75000 j
Change in momentum
p₂ - p₁
75000j - 60000i
Magnitude of change
=
= 96046 Ns.
Explanation:
Given that,
Mass, m = 0.08 kg
Radius of the path, r = 2.7 cm = 0.027 m
The linear acceleration of a yo-yo, a = 5.7 m/s²
We need to find the tension magnitude in the string and the angular acceleration magnitude of the yo‑yo.
(a) Tension :
The net force acting on the string is :
ma=mg-T
T=m(g-a)
Putting all the values,
T = 0.08(9.8-5.7)
= 0.328 N
(b) Angular acceleration,
The relation between the angular and linear acceleration is given by :
(c) Moment of inertia :
The net torque acting on it is, , I is the moment of inertia
Also,
So,
Hence, this is the required solution.
Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:
Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?
Answer:
The Force act on each side is 2.43×10⁴N
Explanation:
Given data
n=3 mol
L=0.3 m
Temperature=20.0°C=293 K
To find
Force F
Solution
To get force act on each side it would employ by
F=P.A
Where P is pressure
A is Area
First we need to find pressure by applying ideal gas law
So
So The Force is given as:
The Force act on each side is 2.43×10⁴N