The two systems that work together to deliver oxygen are D, respiratory and cardiovascular
Answer:
Tension = 0.012 N
Explanation:
If the black widow spider is hanging vertically motionless from the ceiling above. Then, the weight of the spider must be balancing the tension in the spider web. Therefore,
Tension = Weight
Tension = mg
where,
m = mass of spider = 1.27 g = 0.00127 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
Tension = (0.00127 kg)(9.8 m/s²)
<u>Tension = 0.012 N</u>
So the acceleration of gravity is 9.8 m/s so that’s how quickly it will accelerate downwards. You can use a kinematic equation to determine your answer. We know that initial velocity was 19 m/s, final velocity must be 0 m/s because it’s at the very top, and the acceleration is -9.8 m/s. You can then use this equation:
Vf^2=Vo^2+2ax
Plugging in values:
361=19.6x
X=18 m
Answer:
1 cm⁻¹ =1.44K 1 ev = 1.16 10⁴ K
Explanation:
The relationship between temperature and thermal energy is
E = K T
The relationship of the speed of light
c =λ f = f / ν 1/λ= ν
The Planck equation is
E = h f
Let's start the transformations
c = f λ = f / ν
f = c ν
E = h f
E = h c ν
E = KT
h c ν = K T
T = h c ν / K =( h c / K) ν
Let's replace the constants
h = 6.63 10⁻³⁴ J s
c = 3 10⁸ m / s
K = 1.38 10⁻²³ J / K
v = 1 cm-1 (100 cm / 1 m) = 10² m-1
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 1 10²
A = h c / K = 1,441 10⁻²
T = 1.44K
ν = 103 cm⁻¹ = 103 10² m
T = (6.63 10⁻³⁴ 3. 10⁸ / 1.38 10⁻²³) 103 10²
T = 148K
1 Rydberg = 1.097 10 7 m
As we saw at the beginning the λ=1 / v
T = (h c / K) 1 /λ
T = 1,441 10⁻² 1 / 1,097 10⁷
T = 1.3 10⁻⁹ K
E = 1Ev (1.6 10⁻¹⁹ J /1 eV) = 1.6 10⁻¹⁹ J
E = KT
T = E/K
T = 1.6 10⁻¹⁹ /1.38 10⁻²³
T = 1.16 10⁴ K
Answer:

Explanation:
The electrostatic potential energy is given by the following formula

Now, we will apply this formula to both cases:

So, the change in the potential energy is
