C. Three isotopes of the same element because the number of electrons and protons remain the same but the number of neutrons in the nucleus are different. This means that the isotopes have different masses to each other but the same chemical properties.
Answer:
2200000 = 2.2E6 min for light from Proxima to reach earth
8.3 min from light sun to reach earth
2.2E6/8.3 = 2.56E5 times for light from Proxima
Proxima is about 256,000 times farther away than the sun
Since the sun is about 93,000,000 = 9.3E7 miles from earth
Proxima is then 9.3E7 * 2.56E5 = 2.4E13 miles away
Note - the speed of light is
3.00E8 m/s * 60 s/min / 1000 m/km = 1.8E7 km/min as given
Complete Question
A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of 
Answer:
The current is
Explanation:
From the question we are told that
The area is
The initial magnetic field at
is 
The magnetic field at
is 
The resistance is 
Generally the induced emf is mathematically represented as

=> 
=> 
Generally the current induced is mathematically represented as

=>
=>
Answer: because increasing biodiversity can influence ecosystem functions such as productivity and variety and even the likelihood that a particular species is discovered by a comminity
Explanation:
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct