2. Using the solar system data in the

A research group at Dartmouth College has developed a Head Impact Telemetry (HIT) System that can be used to collect data about

Olin [163]

Answer:

6.05 cm

Explanation:

The given equation is

2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)

The initial head velocity V₀ₓ =11 m/s

The final head velocity Vₓ is 0

The accelerationis given by =1000 m/s²

the stopping distance = x-x₀=?

So we can wind the stopping distance by following formula

2 (-1000)(x-x₀)=[ $0^{2} -11^{2}$ ]

x-x₀=6.05* $10^{-2}$ m

=6.05 cm

3 0

2 years ago

A spring is stretched 6 in by a mass that weighs 8 lb. The mass is attached to a dashpot mechanism that has a damping constant o

olya-2409 [2.1K]

Answer:

y= 240/901 cos 2t+ 8/901 sin 2t

Explanation:

To find mass m=weighs/g

m=8/32=0.25

To find the spring constant

Kx=mg (given that c=6 in and mg=8 lb)

K(0.5)=8 (6 in=0.5 ft)

K=16 lb/ft

We know that equation for spring mass system

my''+Cy'+Ky=F

now by putting the values

0.25 y"+0.25 y'+16 y=4 cos 20 t ----(1) (given that C=0.25 lb.s/ft)

Lets assume that at steady state the equation of y will be

y=A cos 2t+ B sin 2t

To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

y'= -2A sin 2t+2B cos 2t

y"=-4A cos 2t-4B sin 2t

Now put the values of y" , y' and y in equation 1

0.25 (-4A cos 2t-4B sin 2t)+0.25(-2A sin 2t+2B cos 2t)+16(A cos 2t+ B sin 2t)=4 cos 20 t

So by comparing the coefficient both sides

30 A+ B=8

A-30 B=0

So we get

A=240/901 and B=8/901

So the steady state response

y= 240/901 cos 2t+ 8/901 sin 2t

6 0

2 years ago

In a distant solar system, a giant planet has

sergeinik [125]

Answer:

mass of the planet: $5.9\,10^{26}\,kg$

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}$

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

$m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}$

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R ( $5.32\,10^5\,km=5.32\,10^8\,m$ , the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference ( $2\,\pi\,R$ ) in 388800 seconds, therefore its tangential velocity is:

$v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}$

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

$v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg$