Let's ask this question step by step:
Part A)
a x b = (3.0i + 5.0j) x (2.0i + 4.0j) = (12-10) k = 2k
ab = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
Part (c)
(a + b) b = [(3.0i + 5.0j) + (2.0i + 4.0j)]. (2.0i + 4.0j)
(a + b) b = (5.0i + 9.0j). (2.0i + 4.0j)
(a + b) b = 10 + 36
(a + b) b = 46
Part (d)
comp (ba) = (a.b) / lbl
a.b = (3.0i + 5.0j). (2.0i + 4.0j) = 6 + 20 = 26
lbl = root ((2.0) ^ 2 + (4.0) ^ 2) = root (20)
comp (ba) = 26 / root (20)
answer
2k
26
46
26 / root (20)
Answer:
R₁ = 50.77 Ω
Explanation:
Since, we know that:
Electric Power = P = VI
but from Ohm's Law:
V = IR
(or) I = V/R
Therefore,
P = V²/R
(OR) R = V²/P
where,
V = Battery Voltage
R = Resistance of combination
FOR SERIES COMBINATION:
R = Rs = (57 V)²/48 W
Rs = 67.69 Ω
but, we know that:
Rs = R₁ + R₂
R₁ + R₂ = 67.69 Ω
R₁ = 67.69 Ω - R₂ __________ eqn (1)
FOR PARALLEL COMBINATION:
R = Rp = (57 V)²/256 W
Rp = 12.69 Ω
but, we know that:
Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω
using eqn (1) and value of R₁ + R₂, we get
Rp = 12.69 = R₂(67.69 - R₂)/67.69
859.08 = 67.69 R₂ - R₂²
R₂² - 67.69 R₂ + 859.08 = 0
Solving this quadratic equation we get the answers:
Either, R₂ = 50.76 Ω
Either, R₂ = 16.92 Ω
Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,
<u>R₂ = 16.92 Ω</u>
using this value in eqn (1), we get:
R₁ = 67.69 Ω - 16.92 Ω
<u>R₁ = 50.77 Ω</u>
Answer:
B) Power is the rate at which work is done
Answer:5.7m/s
Explanation:
Mass=1kg
Initial velocity=u=8m/s
height=h=1.6m
Final velocity =v
Acceleration due to gravity=g=9.8m/s^2
v^2=u^2-2xgxh
v^2=8^2-2x9.8x1.6
v^2=8x8-2x9.8x1.6
v^2=64-31.36
v^2=32.64
Take the square root of both sides
√(v^2)=√(32.64)
v=5.7
Speed at the height of 1.6m is 5.7m/s
Answer:
The earthquake occurred at a distance of 1122 km
Explanation:
Given;
speed of the P wave, v₁ = 8.5 km/s
speed of the S wave, v₂ = 5.5 km/s
The distance traveled by both waves is the same and it is given as;
Δx = v₁t₁ = v₂t₂
let the time taken by the wave with greater speed = t₁
then, the time taken by the wave with smaller speed, t₂ = t₁ + 1.2 min, since it is slower.
v₁t₁ = v₂t₂
v₁t₁ = v₂(t₁ + 1.2 min)
v₁t₁ = v₂(t₁ + 72 s)
v₁t₁ = v₂t₁ + 72v₂
v₁t₁ - v₂t₁ = 72v₂
t₁(v₁ - v₂) = 72v₂
The distance traveled is given by;
Δx = v₁t₁
Δx = (8.5)(132)
Δx = 1122 km
Therefore, the earthquake occurred at a distance of 1122 km