Answer:
Speed of the helium after collision = 246 m/s
Explanation:
Given that
Mass of helium ,m₁ = 4 u
u₁=598 m/s
Mass of oxygen ,m₂ = 32 u
u₂ = 401 m/s
v₂ =445 m/s
Given that initially both are moving in the same direction and lets take they are moving in the right direction.
Speed of the helium after collision = v₁
There is no any external force on the masses that is why the linear momentum will be conserve.
Initial linear momentum = Final linear momentum
P = m v
m₁u₁+m₂u₂ = m₁v₁+m₂v₂
598 x 4 + 32 x 401 = 4 x v₁+ 32 x 445
v₁ = 246 m/s
Speed of the helium after collision = 246 m/s
The spring constant is 4 N/m
Explanation:
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law:

where
F is the magnitude of the spring applied
k is the spring constant
x is the elongation of the spring, relative to its equilibrium position
For the spring in this problem, we have:
F = 0.12 N (force applied)
x = 3 cm = 0.03 m (elongation of the spring)
Therefore, we can solve the formula for k to find the spring constant:

Learn more about forces:
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<span>Avogadro's number
represents the number of units in one mole of any substance. This has the value
of 6.022 x 10^23 units / mole. This number can be used to convert the number of
atoms or molecules into number of moles.
</span>2.39 moles Ne ( 6.022 x 10^23 atoms / mole ) = 1.44 × 10^24 atoms Ne
Answer:
(a) q = 2.357 x 10⁻⁵ C
(b) Φ = 2.66 x 10⁶ N.m²/C
Explanation:
Given;
diameter of the sphere, d = 1.1 m
radius of the sphere, r = 1.1 / 2 = 0.55 m
surface charge density, σ = 6.2 µC/m²
(a) Net charge on the sphere
q = 4πr²σ
where;
4πr² is surface area of the sphere
q is the net charge on the sphere
σ is the surface charge density
q = 4π(0.55)²(6.2 x 10⁻⁶)
q = 2.357 x 10⁻⁵ C
(b) the total electric flux leaving the surface of the sphere
Φ = q / ε
where;
Φ is the total electric flux leaving the surface of the sphere
ε is the permittivity of free space
Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)
Φ = 2.66 x 10⁶ N.m²/C
Answer:
The magnetic field in the System is 0.095T
Explanation:
To solve the exercise it is necessary to use the concepts related to Faraday's Law, magnetic flux and ohm's law.
By Faraday's law we know that

Where,
electromotive force
N = Number of loops
B = Magnetic field
A = Area
t= Time
For Ohm's law we now that,
V = IR
Where,
I = Current
R = Resistance
V = Voltage (Same that the electromotive force at this case)
In this system we have that the resistance in series of coil and charge measuring device is given by,

And that the current can be expressed as function of charge and time, then

Equation Faraday's law and Ohm's law we have,



Re-arrange for Magnetic Field B, we have

Our values are given as,





Replacing,


Therefore the magnetic field in the System is 0.095T