Because you know that gravity is in m/s^2 so, period will be measured in seconds. You know the cable is 12m long and gravity is 9.81 solve for T (period) 2π12sqrt(9.81)=6.94922
Answer:
Explanation:
a )
While breaking initial velocity u = 62.5 mph
= 62.5 x 1760 x 3 / (60 x 60 ) ft /s
= 91.66 ft / s
distance trvelled s = 150 ft
v² = u² - 2as
0 = 91.66² - 2 a x 150
a = - 28 ft / s²
b ) While accelerating initial velocity u = 0
distance travelled s = .24 mi
time = 19.3 s
s = ut + 1/2 at²
s is distance travelled in time t with acceleration a ,
.24 = 0 + 1/2 a x 19.3²
a = .001288 mi/s²
= 2.06 m /s²
c )
If distance travelled s = .25 mi
final velocity v = ? a = .001288 mi / s²
v² = u² + 2as
= 0 + 2 x .001288 x .25
= .000644
v = .025 mi / s
= .0025 x 60 x 60 mi / h
= 91.35 mph .
d ) initial velocity u = 59 mph
= 86.53 ft / s
final velocity = 0
acceleration = - 28 ft /s²
v = u - at
0 = 86.53 - 28 t
t = 3 sec approx .
-GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
Given
A particle of mass m moving under the influence of a fixed mass's M, gravitational potential energy of formula -GMm/r, where r is the separation between the masses and G is the gravitational constant of the universe.
As the Gravity Potential energy of particle = -GMm/r
Total energy of particle = Kinetic energy + Potential Energy
As we know that
Kinetic energy = 1/2mv²
Also, v is equals to square root of GM/r
v = √GM/r
Put the value of v in the formula of kinetic energy
We get,
Kinetic Energy = GMm/2r
Total Energy = GMm/2r + (-GMm/r)
= GMm/2r - GMm/r
= -GMm/2r
Hence, -GMm/2r is the total energy of the mass m if it is in a circular orbit about mass M.
Learn more about Gravitational Potential Energy here brainly.com/question/15896499
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To solve this problem we will apply the laws of Mersenne. Mersenne's laws are laws describing the frequency of oscillation of a stretched string or monochord, useful in musical tuning and musical instrument construction. This law tells us that the velocity in a string is directly proportional to the root of the applied tension, and inversely proportional to the root of the linear density, that is,
Here,
v = Velocity
= Linear density (Mass per unit length)
T = Tension
Rearranging to find the Period we have that
As we know that speed is equivalent to displacement in a unit of time, we will have to
Therefore the tension is 5.54N
