Answer:
The farther star will appear 4 times fainter than the star that is near to the observer.
Explanation:
Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time
Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)
This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by
For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by
Hence we sense it as 4 times fainter than the nearer star.
<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16
Traveling against the current:
(18 - c)*t = 8
Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16
Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16
(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0
Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c
So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3
(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3
And it's verified.</span>
Answer:
the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Explanation:
a) Kinetic energy of block = potential energy in spring
½ mv² = ½ kx²
Here m stands for combined mass (block + bullet),
which is just 1 kg. Spring constant k is unknown, but you can find it from given data:
k = 0.75 N / 0.25 cm
= 3 N/cm, or 300 N/m.
From the energy equation above, solve for v,
v = v √(k/m)
= 0.15 √(300/1)
= 2.598 m/s.
b) Momentum before impact = momentum after impact.
Since m = 1 kg,
v = 2.598 m/s,
p = 2.598 kg m/s.
This is the same momentum carried by bullet as it strikes the block. Therefore, if u is bullet speed,
u = 2.598 kg m/s / 8 × 10⁻³ kg
= 324.76 m/s.
Hence, the magnitude of the velocity of the block just after impact is 2.598 m/s and the original speed of the bullect is 324.76m/s.
Answer:
442.5 rad
Explanation:
w₀ = initial angular velocity of the disk = 7.0 rad/s
α = Constant angular acceleration = 3.0 rad/s²
t = time period of rotation of the disk = 15 s
θ = angular displacement of the point on the rim
Angular displacement of the point on the rim is given as
θ = w₀ t + (0.5) α t²
inserting the values
θ = (7.0) (15) + (0.5) (3.0) (15)²
θ = 442.5 rad
The answer is Concave lenses
