You asked a question. I'm about to answer it.
Sadly, I can almost guarantee that you won't understand the solution.
This realization grieves me, but there is little I can do to change it.
My explanation will be the best of which I'm capable.
Here are the Physics facts I'll use in the solution:
-- "Apparent magnitude" means how bright the star appears to us.
-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).
-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by ⁵√100 .
That's about 2.512... .
-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .
-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).
That's all the Physics. The rest of the solution is just arithmetic.
____________________________________________________
-- The star in the question would appear M(-5) at a distance of
32.6 light years.
-- It actually appears as a M(+5). That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.
-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .
-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is
(32.6) · (100)^(1) light years
= (32.6) · (100) light years
= approx. 3,260 light years . (roughly 1,000 parsecs)
I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
Answer:
W=561.41 J
Explanation:
Given that
m = 51 kg
μk = 0.12
θ = 36.9∘
Lets F is the force applied by man
Given that block is moving at constant speed it mans that acceleration is zero.
Horizontal force = F cos θ
Vertical force = F sinθ
Friction force Fr= μk N
N + F sinθ = m g
N = m g - F sinθ
Fr = μk (m g - F sinθ)
For equilibrium
F cos θ = μk (m g - F sinθ)
F ( cos θ +μk sinθ) = μk (m g
Now by putting the values
F ( cos 36.9∘ + 0.12 x sin36.9∘)=0.12 x 51 x 10
F= 70.2 N
We know that Work
W= F cos θ .d
W= 70.2 x cos 36.9∘ x 10
W=561.41 J
acceleration is considered to describe an increase or positive change of speed or velocity But deceleation is considered to describe a decrease or negative change of speed or velocity
K.E = 0.5 * m * v^2
= 0.5 * 80 * (1.5)^2
K.E = 90 joule
This question is incomplete, the complete question is;
An l-c cirucit with a 70 mh inductor and a 0.54 μF capacitor oscillates. The maximum charge on the capacitor is 11.5 μC. What are the oscillation frequency and the maximum current in this circuit
;
Options
a) 1.07 kHz, 63.4 mA
b) 4.38 kHz, 101.3 mA
c) 6.74 kHz, 55.7 mA
d) 2.31 kHz, 93.5 mA
e) 0.82 kHz, 59.1 mA
Answer:
the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.
so Option e) 0.82 kHz, 59.1 mA is the correct answer
Explanation:
Given that;
inductor L = 70 mH = 70 × 10⁻³ H
Capacitor C = 0.54 μf = 0.54 × 10⁻⁶ f
Qmax on capacitor = 11.5 μf = 11.5 × 10⁻⁶ c
oscillation frequency in L-C circuit;
f = 1/2π√(LC)
we substitute our values;
f = 1/2π√(70 × 10⁻³ × 0.54 × 10⁻⁶ )
f = 0.0818 × 10⁴ Hz
f = 0.082 × 10³ Hz ≈ 0.82 kHz
Maximum circuit in L-C circuit is given by
I_max = Qmax/√(LC)
we substitute
I_max = 11.5 × 10⁻⁶ / √(70 × 10⁻³ × 0.54 × 10⁻⁶ )
= 59.1 × 10³ A ≈ 59.1 mA
Therefore the oscillation frequency and the maximum current in this circuit are; 0.82 kHz and 59.1 mA respectively.
so Option e) 0.82 kHz, 59.1 mA is the correct answer
