Answer:11.59 J
Explanation:
Given
mass of Particle 
Initially Particle moves towards left 
Final velocity of Particle is towards Right 
According to Work Energy theorem
Work done by all the Forces=change in Kinetic Energy
Work done by Force
![W=\frac{69\times 10^{-3}}{2}\left [ 31^-25^2\right ]](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B69%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5Cleft%20%5B%2031%5E-25%5E2%5Cright%20%5D)
![W=\frac{69\times 10^{-3}}{2}\left [ 961-625\right ]](https://tex.z-dn.net/?f=W%3D%5Cfrac%7B69%5Ctimes%2010%5E%7B-3%7D%7D%7B2%7D%5Cleft%20%5B%20961-625%5Cright%20%5D)
Answer:
The correct answer is a. Both are the same
Explanation:
For this calculation we must use the gravitational attraction equation
F = G m M / r²
Where M will use the mass of the Earth, m the mass of the girl and r is the distance of the girl to the center of the earth that we consider spherical
To better visualize things, let's repair the equation a little
F = m (G M / r²)
The amount in parentheses called acceleration of gravity, entered the force called peos
g = G M / r²
F = W
W = m g
When analyzing this equation we see that the variation in the weight of the girl depends on the distance, which is the radius of the earth plus the height where the girl is
r = Re + h
Re = 6.37 10⁶ m
r² = (Re + h)²
r² = Re² (1 + h / Re)²
Let's replace
W = m (GM / Re²) (1+ h / Re)⁻²
W = m g (1+ h / Re)⁻²
This is the exact expression for weight change with height, but let's look at its values for some reasonable heights h = 6300 m (very high mountain)
h / Re = 10
⁻³
(1+ h / Re)⁻² = 0.999⁻²
Therefore, the negligible weight reduction, therefore, for practical purposes the weight does not change with the height of the mountain on Earth
The correct answer is a
Answer:
5.90 ft/s^2
Explanation:
There are mixed units in this question....convert everything to miles or feet
and hr to s
28 mi / hr = 41.066 ft/s
Displacement = vo t + 1/2 at^2
599 = 41.066 (8.9) + 1/2 a (8.9^2)
solve for a = ~ 5.90 ft/s^2
Due to the law of conservation of momentum, the force exerted on the mallet is equal and opposite to the force exerted on the ball, so the answer is C.
