Answer:
The radius of the loop is 20.66 km
Explanation:
let the radius of the loop be r
mass of airplane is m
At the top, the pilot experiences two radial forces, which are
1) Gravitational force is mg
2) Centrifugal forces mv²/r out of the center
When the pilot experiences no weight,
then, mg = mv²/r
r = v² / g
= 450² / 9.8
= 20.66 x 10³3
= 20.66 km
Answer:
33.14 m/s
Explanation:
The mass of the block is 121g or .121 kg. As the bullet is lodged in the block the total mass is 121+14 = 135 g or 0.135 kg.
The frictional force that makes the block come to a stop is normal force* coefficient of friction = 0.135 * 9.8 * 0.7 = 0.9261 N
As the block comes to rest after sliding for 8.3 meters the energy it was given by the bullet is
0.135 * 9.8 * 0.7 * 8.3
= 7.69 Nm
Now this energy is provided the bullet. So the energy in the bullet was equal to
1/2 * mv² = 0.5 * 14 * v².
0.5 * 0.014 * v^2 = 0.135 * 9.8 * 0.7 * 8.3 = 7.69
=> 0.007 * v² = 7.69
=> v² = 7.69 / 0.007
=> v² = 1098.57
=> v = √1098.57
=> v = 33.14 m/s
Answer:
Explanation:
Lets take
Resistance of bulb 1 =R₁
Resistance of bulb 2 =R₂
As we know that power P
P= ΔV²/R
Given that voltage difference is same for both bulbs
So
P₁R₁= ΔV² --------1
P₂R₂= ΔV² -----------2
When these resistance are connected in series then equivalent resistance R
R=R₁+ R₂
The new power P'
P'=ΔV²/R
P'R=ΔV² ------3
From equation 1 ,2 and 3
P'(R₁+ R₂) = ΔV²