Answer:
D). energy resulting from the attraction between two masses.
Explanation:
The answer for this is b 3.500.000j
No. You cannot rule out the battery even after the open circuit voltage measurement. The open-circuit voltage may not have changed but the battery's internal resistance may have greatly increased.
Answer:
1) t=1.743 sec
2)Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)Vf=17.08 m/s
Explanation:
1)From second equation of motion we get
h=Vit+(1/2)gt^2
here in case(a): Vi=0 m/s,h=14.9m,,put these values in above equation to find the time the ball is in motion
14.9=(0)*t+(1/2)(9.8)t^2
t^2=14.9/4.9
t^2=3.040 sec
t=1.743 sec
2) s=Vo*t
Putting values we get
107=Vo*1.743
Vo=61.388 m/sec
3)the x component of its velocity just be- fore it strikes the ground is the same as the initial velocity of the ball that is=61.388 m/sec
4)From third equation of motion we know that
Vf^2-Vi^2=2gh
here Vi=0 m/s,h=14.9 m
Vf^2=Vi^2+2gh=0+2(9.8)(14.9)
Vf^2=292.04
Vf=17.08 m/s
Answer:
Radius of the loop is 0.18 m or 18 cm
Explanation:
Given :
Current flowing through the wire, I = 45 A
Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T
Number of turns in circular wire, N = 1
Consider R be the radius of the circular wire.
The magnetic field at the center of the current carrying circular wire is determine by the relation:
Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.
Substitute the suitable values in the above equation.

R = 0.18 m