Spacecraft used is "Friendship 7". Hope it helps.
Answer:
6.6 atm
Explanation:
Using the general gas law
P₁V₁/T₁ = P₂V₂/T₂
Let P₂ be the new pressure
So, P₂ = P₁V₁T₂/V₂T₁
Since V₂ = 2V₁ , P₁ = 12 atm and T₁ = 273 + t where t = temperature in Celsius
T₂ = 273 + 2t (since its Celsius temperature doubles).
Substituting these values into the equation for P₂, we have
P₂ = P₁V₁(273 + 2t)/2V₁(273 + t)
P₂ = 12(273 + 2t)/[2(273 + t)]
P₂ = 6(273 + 2t)/(273 + t)]
assume t = 30 °C on a comfortable spring day
P₂ = 6(273 + 2(30))/(273 + 30)]
P₂ = 6(273 + 60))/(273 + 30)]
P₂ = 6(333))/(303)]
P₂ = 6.6 atm
We are given
E = <span>2.64 × 10-21 J
h = </span><span>6.6 × 10-34 J s
The options given below are frequencies, therefore, the question must be asking about the frequency fo the given wave
The equation is
E = h f
Simply substitute and solve for f which is the frequency
f = </span>2.64 × 10-21 J / 6.6 × 10-34 J s
f = <span>4.00 × 1012<span> hertz</span></span>
The force of the push and acceleration of the cart are related such as The unbalanced push causes the cart to accelerate.
Answer:
The rate of change of distance between the two ships is 18.63 km/h
Explanation:
Given;
distance between the two ships, d = 140 km
speed of ship A = 30 km/h
speed of ship B = 25 km/h
between noon (12 pm) to 4 pm = 4 hours
The displacement of ship A at 4pm = 140 km - (30 km/h x 4h) =
140 km - 120 km = 20 km
(the subtraction is because A is moving away from the initial position and the distance between the two ships is decreasing)
The displacement of ship B at 4pm = 25 km/h x 4h = 100 km
Using Pythagoras theorem, the resultant displacement of the two ships at 4pm is calculated as;
r² = a² + b²
r² = 20² + 100²
r = √10,400
r = 101.98 km
The rate of change of this distance is calculated as;
r² = a² + b²
r = 101.98 km, a = 20 km, b = 100 km
