initial velocity of the car given as
final velocity is given as
as we know that
now we can convert final speed into m/s
now acceleration is rate of change in velocity
so the acceleration of the car is 3 m/s^2
Ill provide the answer choices here, assuming its from edge.
A) Sasha’s monthly expenses would be less for buying than for renting.
B) The extra expenses in the mortgage payment cover all maintenance and repairs.
C) Sasha’s down payment will likely be less if she decided to buy.
D) Sasha will own the house and earn equity as its value increases.
the correct answer is D) Sasha will own the house and earn equity as its value increases.
Answer:
k = 9.6 x 10^5 N/m or 9.6 kN/m
Explanation:
First, we need to use the expression to calculate the spring constant which is:
w² = k/m
Solving for k:
k = w²*m
To get the angular velocity:
w = 2πf
The problem is giving the linear velocity of the car which is 5.7 m/s. With this we can calculate the frequency of the car:
f = V/x
f = 5.7 / 4.9 = 1.16 Hz
Now the angular velocity:
w = 2π*1.16
w = 7.29 rad/s
Finally, solving for k:
k = (7.29)² * 1800
k = 95,659.38 N/m
In two significant figures it'll ve 9.6 kN/m
Answer: A:The reproductive system produces hormones.
B:The reproductive system transports reproductive cells.
C:The reproductive system produces reproductive cells.
Explanation:
A:The reproductive system produces hormones. : The reproductive organs ovaries in females and testes in males produces hormones. The ovaries produce estrogen and progesterone and testes produce testosterone.
B:The reproductive system transports reproductive cells. : The ovaries in females transfers an egg into the fallopian tube and testes in males secrete sperms at the time of copulation the fertilization of egg and sperm leads to the development of zygote the precursor of new life.
C:The reproductive system produces reproductive cells.: The ovaries undergo oogenesis and in testes spermatogenesis takes place to develop egg and sperms respectively.
Answer:
x(t)=0.337sin((5.929t)
Explanation:
A frictionless spring with a 3-kg mass can be held stretched 1.6 meters beyond its natural length by a force of 90 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 2 m/sec, find the position of the mass after t seconds.
Solution. Let x(t) denote the position of the mass at time t. Then x satisfies the differential equation
Definition of parameters
m=mass 3kg
k=force constant
e=extension ,m
ω =angular frequency
k=90/1.6=56.25N/m
ω^2=k/m= 56.25/1.6
ω^2=35.15625
ω=5.929
General solution will be
differentiating x(t)
dx(t)=-5.929c1sin(5.929t)+5.929c2cos(5.929t)
when x(0)=0, gives c1=0
dx(t0)=2m/s gives c2=0.337
Therefore, the position of the mass after t seconds is
x(t)=0.337sin((5.929t)
