Answer:
The ratio of
and
is 0.9754
Explanation:
Given that,
Distance z = 4.50 d
First equation is


Second equation is

We need to calculate the ratio of
and 
Using formula


Put the value into the formula


Hence, The ratio of
and
is 0.9754
Answer:
work done on the object by gravitational force = 0 joules
Explanation:
work = force × displacement
force = mass × acceleration
so,
work = mass × acceleration × displacement
we know that mass= 10 kg , gravitational acceleration= 9.8 
and displacement= 0 m since the object is not moving vertically.
so,
work = 10 × 9.8 × 0 = 0 joules
Answer:
(a) The least time required for the rotation is 50 seconds
(b) The corresponding value of α₁ is 1.6 rad/s²
Explanation: Please see the attachments below
Answer:
N₁ = 393.96 N and N = 197.96 N
Explanation:
In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop
Lowest point, we write Newton's second law n for the y-axis
N -W = m a
where the acceleration is ccentripeta
a = v² / r
N = W + m v² / r
N = mg + mv² / r
we can use energy to find the speed at the bottom of the circle
starting point. Highest point where the ball is released
Em₀ = U = m g h
lowest point. Stop curl down
= K = ½ m v²
Emo = Em_{f}
m g h = ½ m v²
v² = 2 gh
we substitute
N = m (g + 2gh / r)
N = mg (1 + 2h / r)
let's calculate
N₁ = 5 9.8 (1 + 2 17.6 / 5)
N₁ = 393.96 N
headed up
we repeat the calculation in the longest part of the loop
-N -W = - m v₂² / r
N = m v₂² / r - W
N = m (v₂²/r - g)
we seek speed with the conservation of energy
Em₀ = U = m g h
final point. Top of circle with height 2r
= K + U = ½ m v₂² + mg (2r)
Em₀ = Em_{f}
mgh = ½ m v₂² + 2mgr
v₂² = 2 g (h-2r)
we substitute
N = m (2g (h-2r) / r - g)
N = mg (2 (h-r) / r 1) = mg (2h/r -2 -1)
N = mg (2h/r - 3)
N = 5 9.8 (2 17.6 / 5 -3)
N = 197.96 N
Directed down
Answer:
Explanation:
Electric field at the surface of the the lead 208 = KQ/ R²
where K = 8.99 × 10⁹ Nm² /C²
Q ( total charge inside the nucleus) and e is the charge of a proton = Ne = 82 × 1.6 × 10⁻¹⁹ C = 1.312 × 10⁻¹⁷ C
V of the lead = 208 v of a proton assuming they both are sphere
4/3 πR³ =208( 4/3 πr³) where R is the radius of the sphere and r is the radius of the proton
R³ = 208 r³
R = ∛( 208 r³) = 5.92r
replace r with 1.20 x 10-15 m
R = 5.92 ×1.20 x 10-15 m = 7.11 × 10⁻¹⁵ m
E = ( 8.99 × 10⁹ Nm² /C² × 1.312 × 10⁻¹⁷ C ) / (7.11 × 10⁻¹⁵ m)² = 0.233 × 10²² N/C = 2.33 × 10²¹ N/C