Answer:
k = 52.2 N / m
Explanation:
For this exercise we are going to use the conservation of mechanical energy.
Starting point. When it is 30 m high
Em₀ = K + U = ½ m v² + m g h
Final point. Right when you hit the water
= K_{e} = ½ k x²
in this case the distance the bungee is stretched is 30 m
x = h
as they indicate that there are no losses, energy is conserved
Em₀ = Em_{f}
½ m v² + m g h = ½ k h²
k = 
let's calculate
k =
k = 52.2 N / m
Answer:
a) V = -0.227 mV
b) V = -0.5169 mV
Explanation:
a)
Inside a sphere with a uniformly distributed charge density, electric field is radial and has a magnitude
E = (qr) / (4πε₀R³)
As we know that
V = -
By solving above equation, we get
V = (-qr²) / (8πε₀R³)
When
R = 1.81 cm
r = 1.2 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-2.80 × 10⁻¹⁵ × (1.2 × 10⁻²)²) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²)³)
V = -2.27 × 10⁻⁴ V
V = -0.227 mV
b)
When
r = R
R = 1.81 cm
q = +2.80 fC
ε₀ = 8.85 × 10⁻¹²
V = (-qR²) / (8πε₀R³)
V = (-q) / (8πε₀R)
V = (-2.80 × 10⁻¹⁵) / (8 × 3.14 × 8.85 × 10⁻¹² × (1.81 × 10⁻²))
V = -5.169 × 10⁻⁴ V
V = -0.5169 mV
The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 × (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
= 8.077 × 10^5 m/s
Answer: 510 m/s
Explanation: specific gravity of steam is 18/29 = 0.620
It is the ratio of the density of steam over density of water
400m3/s of steam =
400m3ms * 0.620 of water
= 248m3/s of water
Total flow rate Q = 248 + 7 = 255m3/s
Using Q = AV
Where A is area and V is velocity
V = Q/A
V = 255/0.5 = 510m/s
<u>Explanation:</u>
In other to avoid a situation where their differences, pride, and personalities caused them to compete and publicly insult one another instead of working together they need to respectfully consider the viewpoints of each other, as well as assign individual roles that should be respected.