Answer:
v = √(10gh/7)
Explanation:
Initial gravitational energy = final rotational energy + kinetic energy
PE = RE + KE
mgh = ½ Iω² + ½ mv²
For a solid sphere, I = ⅖ mr².
For rolling without slipping, ω = v/r.
mgh = ½ (⅖ mr²) (v/r)² + ½ mv²
mgh = ⅕ mv² + ½ mv²
mgh = 7/10 mv²
10/7 gh = v²
v = √(10gh/7)
<span>The Special (as in Limited) Theory of Relativity, is a simplification of the General Theory of Relativity.
Essentially, if you eliminate acceleration, and any significant mass from the General Theory, you get the Special Theory.
Evidence for Special Relativity (solar moons for example), is also evidence for the General Theory. The General Theory is supported by:
- Universal expansion
- the spin down of binary pulsars
- frame dragging
- gravitational lensing
- gravitational time dilation</span><span>
</span>
(a) We will use the equation v = u + at
Initial velocity u = 5.00 m/s
Acceleration a = 0.0600 m/s²
time = 8 min = 8 x 60 = 480 s
Final velocity
= u + at
= 5.00 + 0.0600(480)
= 33.8 m/s
The final velocity is 33.8 m/s
because of conduction the metal spoon is a conductor so the heat is getting traveled through the spoon and into your hand
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>
