Answer: B. Mechanical Waves
Explanation:
U = 0, initial upward speed
a = 29.4 m/s², acceleration up to 3.98 s
a = -9.8 m/s², acceleration after 3.98s
Let h₁ = the height at time t, for t ≤ 3.98 s
Let h₂ = the height at time t > 3.98 s
Motion for t ≤ 3.98 s:
h₁ = (1/2)*(29.4 m/s²)*(3.98 s)² = 232.854 m
Calculate the upward velocity at t = 3.98 s
v₁ = (29.4 m/s²)*(3.98 s) = 117.012 m/s
Motion for t > 3.98 s
At maximum height, the upward velocity is zero.
Calculate the extra distance traveled before the velocity is zero.
(117.012 m/s)² + 2*(-9.8 m/s²)*(h₂ m) = 0
h₂ = 698.562 m
The total height is
h₁ + h₂ = 232.854 + 698.562 = 931.416 m
Answer: 931.4 m (nearest tenth)
Answer:
(a) The current in the wire is 19.89 A
(b) The distance from the wire is 0.159 cm
Explanation:
Given;
magnetic field, B = 2.5 mT
diameter of the wire, d = 1 cm
radius of the wire, r = 0.5 cm = 0.005 m
(a) The current in the wire is calculated as;
(b) The distance from the wire where the magnetic field is 2.5 mT is calculated as;
Given:
m₁ = 1540 g, mass of iron horseshoe
T₁ = 1445 °C, initial temperature of horseshoe
c₁ = 0.4494 J/(g-°C), specific heat
m₂ = 4280 g, mass of water
T₂ = 23.1 C, initial temperature of water
c₂ = 4.18 J/(g-°C), specific heat of water
L = 947,000 J heat absorbed by the water.
Let the final temperature be T °C.
For energy balance,
m₁c₁(T₁ - T) = m₂c₂(T - T₂) + L
(1540 g)*(0.4494 J/(g-C))*(1445-T C) = (4280 g)*(4.18 J/(g-C))*(T-23.1 C) + 947000 J
692.076(1445 - T) = 17890(T - 23.1) + 947000
10⁶ - 692.076T = 17890T - 413259 + 947000
466259 = 18582.076T
T = 25.09 °C
Answer: 25.1 °C
Answer:
92.397amu
Explanation: The exact amu of the mystery element is obtained by multiplying the relative abundance of each individual isotope by its respective amu and then summing the results.
The sum of the total relative abundance for all the isotopes should be 100%.
However, the relative abundance of the isotope with 95.502amu is not given; therefore to obtain it we subtract the sum of the known relative abundances from 100% as follows:
Relative abundance of isotope with 95.502amu = 100-(23.63+30.53) = 42.84%