Answer:
v = 10 m/s
Explanation:
Given that,
Distance covered by a sprinter, d = 100 m
Time taken by him to reach the finish line, t = 10 s
We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,
v = d/t
Hence, his average velocity is 10 m/s.
Answer:
The transverse component of acceleration is 26.32 
where as radial the component of acceleration is 8.77
Explanation:
As per the given data
u=π/4 rad
ω=u'=2 rad/s
α=u''=4 rad/s
So the transverse component of acceleration are given as
Here
So
The transverse component of acceleration is 26.32
The radial component is given as
Here
So
The radial component of acceleration is 8.77
Answer:
Which of the following was not a reaction to John Brown's execution because of his failed raid on Harper's Ferry ???????
Explanation:
Answer:
0.08kg
Explanation:
K.E = 1/2 mv^2
v = 970m/s
K.E = 3.9x 10^3J= 3900J
K.E = 1/2 mv^2
3900 = 1/2 m x 970x 970
3900 = 1/2 ×940900m
3900 = 470450m
m = 3900/470450 = 0.00828993516 = 0.008kg
Answer:
Concepts and Principles
1- Kinetic Energy: The kinetic energy of an object is:
K=1/2*m*v^2 (1)
where m is the object's mass and v is its speed relative to the chosen coordinate system.
2- Gravitational potential energy of a system consisting of Earth and any object is:
U_g = -Gm_E*m_o/r*E-o (2)
where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.
Solution
The argument:
My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.
The counterargument:
We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):
K=1/2*m*v^2 (1)
and a gravitational potential energy Ug given by Equation (2):
Ug = -G*Mm/R
where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:
K+U=0
1/2*m*v^2 + (-G*Mm/R) = 0
1/2*m*v^2 = G*Mm/R
1/2*v^2 = G*M/R
solving for v we get
v = √2G*M/R
so we see v does not depend on the mass of the projectile
