Answer:
As b ∝ (L/r²) and
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
hence,
the Sun appear brighter in the sky
Explanation:
The brightness (b) is directly proportional to the Luminosity of the star (L) and inversely proportional to the square of the distance between the star and the observer (r).
thus, mathematically,
b ∝ (L/r²)
now,
given
L for sirius is 23 times more than the sun i.e 23L
now,
the distance of the sun from the earth is 0.00001581 light years
and
the distance of the Sirius from the earth is 8.6 light years
thus,
using the the relation between conclude that the value of brightness for the Sirius comes very very low as compared to the value for brightness for the Sun.
hence, the sun appears brighter
Answer:
u = - 38.85 m/s^-1
Explanation:
given data:
acceleration = 2.10*10^4 m/s^2
time = 1.85*10^{-3} s
final velocity = 0 m/s
from equation of motion we have following relation
v = u +at
0 = u + 2.10*10^4 *1.85*10^{-3}
0 = u + (21 *1.85)
0 = u + 38.85
u = - 38.85 m/s^-1
negative sign indicate that the ball bounce in opposite directon
Answer:
hope this will help you
have a great day
Explanation:
please mark me as the brainliest
250/4= 62.5 mph
to find the mph of a car, you need to divide the number of miles traveled by the hours that it took to travel that many miles
Answer:
576 joules
Explanation:
From the question we are given the following:
weight = 810 N
radius (r) = 1.6 m
horizontal force (F) = 55 N
time (t) = 4 s
acceleration due to gravity (g) = 9.8 m/s^{2}
K.E = 0.5 x MI x ω^{2}
where MI is the moment of inertia and ω is the angular velocity
MI = 0.5 x m x r^2
mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg
MI = 0.5 x 82.65 x 1.6^{2}
MI = 105.8 kg.m^{2}
angular velocity (ω) = a x t
angular acceleration (a) = torque ÷ MI
where torque = F x r = 55 x 1.6 = 88 N.m
a= 88 ÷ 105.8 = 0.83 rad /s^{2}
therefore
angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s
K.E = 0.5 x MI x ω^{2}
K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules