Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Answer:
Explanation:
An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.
Answer:
m = 2301.8 [kg]
Explanation:
This problem can be solved in an easy way using the definition of kinetic energy, in this way you can clear the mass, depending on the known variables.
![E_{k}=0.5*m*v^2\\replacing\\\\9023 = 0.5 *m*(2.8)^2\\m = 2301.8[kg]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E2%5C%5Creplacing%5C%5C%5C%5C9023%20%3D%200.5%20%2Am%2A%282.8%29%5E2%5C%5Cm%20%3D%202301.8%5Bkg%5D)
So the force that the gymnast with a mass m=45 kg, has to exert against the ground to stop if her acceleration is a=8*g, where g=9.81 m/s², can be obtained from the Newtons second law: F=m*a, where F is the force, m is the mass and a is acceleration.
F=m*a=m*8*g=45*8*9.81=3531.6 N.
So the force that a gymnast has to exert on the mat in order to stop is F=3531.6 N.
The impulse experienced by an object is the force. • time.
The momentum change of an object is the mass. • velocity change.
The impulse equals the momentum change.
Your impulse is 1.875 I hope this helps
