Using the Equation:
v² = vi² + 2 · a · s → Eq.1
where,
v = final velocity
vi = initial velocity
a = acceleration
s = distance
<span><span>We know that vi = 0 because the ball was at rest initially.
</span><span>
Therefore,
Solving Eq.1 for acceleration,
</span></span> v² = vi² + 2 · a · s
v² = 0 + 2 · a · s
v² = 2 · a · s
Rearranging for a,
a = v ²/2·<span>s
Substituting the values,
a = 46</span>²/2×1<span>
a = 1058 m/s</span>²
<span>Now applying Newton's 2nd law of motion,
</span>
<span>F = ma
= 0.145</span>×<span>1058
F = 153.4 N</span>
Answer:
Total energy is constant
Explanation:
The laws of thermodynamics state that thermal energy (heat) is always transferred from a hot body (higher temperature) to a cold body (lower temperature).
This is because in a hot body, the molecules on average have more kinetic energy (they move faster), so by colliding with the molecules of the cold body, they transfer part of their energy to them. So, the temperature of the hot body decreases, while the temperature of the cold body increases.
This process ends when the two bodies reach the same temperature: we talk about thermal equilibrium.
In this problem therefore, this means that the thermal energy is transferred from the hot water to the cold water.
However, the law of conservation of energy states that the total energy of an isolated system is constant: therefore here, if we consider the hot water + cold water as an isolated system (no exchange of energy with the surroundings), this means that their total energy remains constant.
Explanation:
the inability to change the position from rest to motion or motion to rest by themselves is inertia.
Answer:
0.363999909622
Explanation:
F = Force
m = Mass = 15.6 g
C = Drag coefficient
ρ = Density of air = 1.21 kg/m³
A = Surface area = 
v = Terminal velocity = 
s = Displacement = 150 m
Force is given by
F = ma
The drag coefficient is 0.363999909622 (ignoring negative sign)
Complete Question:
A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?
Answer:
The magnitude of the magnetic field = 7.24 μT
Explanation:
Inner radius, a = 4.0 mm = 0.004 m
Outer radius, b = 30 mm = 0.03 m
Radius, r = 12 mm = 0.012 m
let h² = b² - a²
h² = 0.03² - 0.004²
h² = 0.000884
Let d² = r² - a²
d² = 0.012² - 0.004²
d² = 0.000128
Current I = 3A
μ = 4π * 10⁻⁷
The magnitude of the magnetic field is given by:
B = 7.24 * 10⁻⁶T
B = 7.24 μT
