Answer:
1. 100 J
2. 225 J
Explanation:
We'll begin by calculating the mass of the object. This can be obtained as follow:
Velocity (v) = 5 ms¯¹
Kinetic energy (KE) = 25 J
Mass (m) =?
KE = ½mv²
25 = ½ × m × 5²
25 = ½ × m × 25
25 = 25m / 2
Cross multiply
25m = 25 × 2
25m = 50
Divide both side by 25
m = 50 / 25
m = 2 Kg
1. Determination of the kinetic energy when the velocity is doubled.
Mass (m) = 2 Kg
Velocity (v) = double the initial velocity
= 2 × 5 ms¯¹
= 10 ms¯¹
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 2 × 10²
KE = ½ × 2 × 100
KE = 100 J
2. Determination of the kinetic energy when the velocity increased three times.
Mass (m) = 2 Kg
Velocity (v) = three times the initial velocity
= 3 × 5 ms¯¹
= 15 ms¯¹
Kinetic energy (KE) =?
KE = ½mv²
KE = ½ × 2 × 15²
KE = ½ × 2 × 225
KE = 225 J
Answer:
or 11616
Explanation:
Since there are 5280 feet in 1 mile
you do 2.2 × 5280
2.2 × 5280 = 11616
Answer:
We could make a frictionless generator.
Explanation:
This would change life by making the cost for energy next to zero
Answer:
a) n = 9.9 b) E₁₀ = 19.25 eV
Explanation:
Solving the Scrodinger equation for the electronegative box we get
Eₙ = (h² / 8m L²2) n²
where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number
In this case En = 19 eV let us reduce to the SI system
En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J
n = √ (In 8 m L² / h²)
let's calculate
n = √ (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²
n = √ (98) n = 9.9
since n must be an integer, we approximate them to 10
b) We substitute for the calculation of energy
In = (h² / 8mL2² n²
In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹ (1.4 10⁻⁹)² 10²
E₁₀ = 3.08 10⁻¹⁸ J
we reduce eV
E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)
E₁₀ = 1.925 101 eV
E₁₀ = 19.25 eV
the result with significant figures is
E₁₀ = 19.25 eV
Answer:
110 yds
Explanation:
Well if 55 yards is 1/2 of the field then 2 x 55 = 110 yards is total field length
