1) push down on the end of the lever, and 2) 3/4 of the way from the fulcrum
Answer:
x(t) = - 6 cos 2t
Explanation:
Force of spring = - kx
k= spring constant
x= distance traveled by compressing
But force = mass × acceleration
==> Force = m × d²x/dt²
===> md²x/dt² = -kx
==> md²x/dt² + kx=0 ------------------------(1)
Now Again, by Hook's law
Force = -kx
==> 960=-k × 400
==> -k =960 /4 =240 N/m
ignoring -ve sign k= 240 N/m
Put given data in eq (1)
We get
60d²x/dt² + 240x=0
==> d²x/dt² + 4x=0
General solution for this differential eq is;
x(t) = A cos 2t + B sin 2t ------------------------(2)
Now initially
position of mass spring
at time = 0 sec
x (0) = 0 m
initial velocity v= = dx/dt= 6m/s
from (2) we have;
dx/dt= -2Asin 2t +2B cost 2t = v(t) --- (3)
put t =0 and dx/dt = v(0) = -6 we get;
-2A sin 2(0)+2Bcos(0) =-6
==> 2B = -6
B= -3
Putting B = 3 in eq (2) and ignoring first term (because it is not possible to find value of A with given initial conditions) - we get
x(t) = - 6 cos 2t
==>
Answer:
102000 kg
Explanation:
Given:
A total Δν = 15 km/s
first stage mass = 1000 tonnes
specific impulse of liquid rocket = 300 s
Mass flow rate of liquid fuel = 1500 kg/s
specific impulse of solid fuel = 250 s
Mass flow of solid fuel = 200 kg/s
First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds
Now,
Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time
or
Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg
Also,
Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time
or
Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg
Therefore,
The total jettisoned mass flow of the fuel in first stage
= 90000 kg + 12000 kg
= 102000 kg
Answer:
A. speed = 7.14 Km/s
B. distance = 1820.7 Km
Explanation:
Given that: a = 14.0 m/
, t = 8.50 minutes.
But,
t = 8.50 = 8.50 x 60
= 510 seconds
A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;
v = u + at
where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.
u = 0
So that,
v = 14 x 510
= 7140 m/s
The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.
B. the distance traveled can be determined by applying second equation of motion.
s = ut +
a
where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.
u = 0
s =
a
=
x 14 x 
= 7 x 260100
= 1820700 m
The distance that the shuttle has traveled during the given time is 1820.7 Km.
Incomplete question as number of moles and length is missing.So I have assumed 3 moles and length of 0.300 m.So the complete question is here:
Three moles of an ideal gas are in a rigid cubical box with sides of length 0.300 m.What is the force that the gas exerts on each of the six sides of the box when the gas temperature is 20.0∘C?
Answer:
The Force act on each side is 2.43×10⁴N
Explanation:
Given data
n=3 mol
L=0.3 m
Temperature=20.0°C=293 K
To find
Force F
Solution
To get force act on each side it would employ by
F=P.A
Where P is pressure
A is Area
First we need to find pressure by applying ideal gas law
So

So The Force is given as:

The Force act on each side is 2.43×10⁴N