Answer:
ΔE = 8.77 × 10¹¹ J
Explanation:
given,
²¹⁴₈₄Po -----> ²¹⁰₈₂Pb + 42 He
Atomic masses: Pb-210 = 209.98284 amu
Po-214 = 213.99519 amu
He-4 = 4.00260 amu
1 kg = 6.022 × 10²⁶ amu;
NA = 6.022 × 10²³ mol⁻¹
c = 2.99792458 × 10⁸ m/s
energy of molecule using equation
ΔE = Δm c²
Δm is mass difference and c is speed of light
Δm = 209.98284 + 4.00260 - 213.99519
Δm = - 0.00975 amu
1 amu = 1.66 x 10⁻²⁷ kg
- 0.00975 amu = - 0.00975 x 1.66 x 10⁻²⁷ Kg
= -0.016185 x 10⁻²⁷ Kg
total mass = 6.022 × 10²³ x -0.016185 x 10⁻²⁷
= -0.097467 x 10⁻⁴ Kg
ΔE = -(0.097467 x 10⁻⁴) (3 x 10^8)²
ΔE = - 8.77 × 10¹¹
ΔE = 8.77 × 10¹¹ J
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
The height of the diving board is given as

now the speed of the diver is given as

when the diver will jump into the water then his displacement in vertical direction is same as that of height of diving board
So we will have



Part b)

plug in the values in the above equation


Part c)
Horizontal distance moved by the diver is given as



so the distance from the edge of the pool is given as


The solution for this problem is:
(10 x 9.8) = 98.1 m/sec^2 acceleration. Time, to travel 9.4cm or (.094m.), at acceleration of 98m/sec^2
= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle
Frequency = (w/2pi), = 32.3050619/2pi
= 32.3050619/6.28318531
= 5.14 Hz would be the answer
Answer:
Explanation:
Let L be the length of the wire.
velocity of pulse wave v = L / 24.7 x 10⁻³ = 40.48 L m /s
mass per unit length of the wire m = 14.5 x 10⁻⁶ x 10⁻³ / 2 x 10⁻² kg / m
m = 7.25 x 10⁻⁷ kg / m
Tension in the wire = Mg , M is mass hanged from lower end.
= .4 x 9.8
= 3.92 N
expression for velocity of wave in the wire
, T is tension in the wire , m is mass per unit length of wire .
40.48 L = 
1638.63 L² = 3.92 / (7.25 x 10⁻⁷)
L² = 3.92 x 10⁷ / (7.25 x 1638.63 )
L² = 3299.64
L = 57.44 m /s