Answer: 7.07 m/s
Explanation:
Mass of runner = 60 kg runner
Kinetic energy = 1500J
Speed of runner = ?
Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass and speed by which it moves.
Hence, K.E = 1/2 x mass x (speed)^2
1500J = 1/2 x 60kg x (speed)^2
1500J = 30kg x (speed)^2
(speed)^2 = 1500J/30kg
(speed)^2 = 50
To get the value of speed, find the square root of 50
speed = √50
speed = 7.07 m/s
Thus, the runner moves as fast as 7.07 m/s
Because of the location of Mg on the periodic table.
Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Answer:
A) U₀ = ϵ₀AV²/2d
B) U₁ = (ϵ₀AV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
C) U₂ = (kϵ₀AV²)/2d
Explanation:
A) The energy stored in a capacitor is given by (1/2) (CV²)
Energy in the capacitor initially
U₀ = CV²/2
V = voltage across the plates of the capacitor
C = capacitance of the capacitor
But the capacitance of a capacitor depends on the geometry of the capacitor is given by
C = ϵA/d
ϵ = Absolute permissivity of the dielectric material
ϵ = kϵ₀
where k = dielectric constant
ϵ₀ = permissivity of free space/air/vacuum
A = Cross sectional Area of the capacitor
d = separation between the capacitor
If air/vacuum/free space are the dielectric constants,
So, k = 1 and ϵ = ϵ₀
U₀ = CV²/2
Substituting for C
U₀ = ϵ₀AV²/2d
B) Now, for U₁, the new distance between plates, d₁ = 3d
U₁ = ϵ₀AV²/2d₁
U₁ = ϵ₀AV²/(2(3d))
U₁ = (ϵ₀AV²)/6d
This means that the new energy of the capacitor is (1/3) of the initial energy before the increased separation.
C) U₂ = CV²/2
Substituting for C
U₂ = ϵAV²/2d
The dielectric material has a dielectric constant of k
ϵ = kϵ₀
U₂ = (kϵ₀AV²)/2d
Answer:
Continuing to play if you have an injury can make that injury worse. A small stress fracture that might have healed quickly can grow into a more serious, more painful fracture that will take longer to heal. Returning to play too soon after a concussion increases your risk of serious brain injury.
Explanation: Trust me it right bro
