Answer:
a) distance d = 293.36ft
b) acceleration a = 14.67ft/s^2
Explanation:
Acceleration is the change in velocity per unit time.
a = ∆v/t ....1
Given;
Initial velocity vi = 30mph × 5280ft/mile × 1/3600s/h
vi = 44ft/s
Final velocity vf = 70mph × 5280ft/mile × 1/3600s/h
vf = 102.67ft/s
time = 4.0s
From equation 1, acceleration is;
a = ∆v/t = (102.67-44)/4 = 14.67ft/s^2
Distance travelled can be given as;
d = ut + 0.5at^2 .....2
u = 44ft/s
t = 4
a = 14.67ft/s^2
Substituting into the equation 2
d = 44(4) + 0.5(14.67×4^2)
d = 293.36ft
Answer: 
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity 
of the planet P1 with a period 
:
(1)
Where:
is the velocity of planet P1
is the radius of the orbit of planet P1
Finding :
(2)
(3)
(4)
On the other hand, we know the gravitational force 
between the star S with mass 
and the planet P1 with mass 
is:
(5)
Where 
is the Gravitational Constant and its value is 
In addition, the centripetal force 
exerted on the planet is:
(6)
Assuming this system is in equilibrium:
(7)
Substituting (5) and (6) in (7):
(8)
Finding :
(9)
(10)
Finally:
(11) This is the mass of the star S
Answer:
<h2>T(Period) = 1.33s</h2><h2>f(Frequency) = 0.75Hz (cycles/second)</h2>
Explanation:
<h2>Given:</h2><h2 /><h2>λ = 4.0m</h2><h2>Amplitude = 25m</h2><h2>d = 24m</h2><h2>s = 8.0s</h2><h2 /><h2>Required:</h2><h2>f = ?</h2><h2>T = ?</h2><h2 /><h2>Analysis:</h2><h2>v = λf</h2><h2>f =N/t</h2><h2>T = 1/f</h2><h2 /><h2>v = d/t</h2><h2 /><h2>Solve:</h2><h2>v = d/t = 24/8.0 → v = 3.0m/s</h2><h2>v =λf → f = v/λ = 3.0/4.0 → f = 0.75Hz</h2><h2>T = 1/f = 1/0.75 → T = 1.33s</h2><h2 /><h2>Hopes this helps. Mark as brainlest plz!</h2>
Very efficient because it uses less then half to do the action then the vacuum actually stores.
