Answer:
They are able to balance torques due to gravity.
Explanation:
When two friends of different masses will balance themselves on see saw then at equilibrium position the see saw will remain horizontal
This condition will be torque equilibrium position where the see saw will not rotate
Here we can say
here we know that force is due to weight of two friends
and their positions are different with respect to the lever about which see saw is rotating
since both friends are of different weight so they will balance themselves are different positions as per above equation
<span>electric, solar, wind, and geothermal.</span>
There are 8 electrons in the third energy level of Calcium atom.
Answer:
F = −10093.41 N
Explanation:
Given that,
Mass of a baseball, m = 143 g = 0.143 kg
Initial speed of the baseball, u = +38.8 m/s
The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s
We need to find the average force exerted on the ball by the bat. So, Force is given by :
a is acceleration
So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.
Answer:
a = 1.428 [m/s²]
v₀ = 5 [m/s]
Explanation:
To solve this problem we must use the following equation of kinematics.
where:
x = final point [m]
x₀ = initial point [m]
v₀ = initial velocity [m/s]
a = acceleration [m/s²]
t = time [s]
But we need to use this additional equation.
where:
vf = final velocity = 15 [m/s]
Now we can use this equation, replacing it, in the first one. We must bear in mind that the difference among x - x₀ is equal to 70 [m]
![x-x_{o}=v_{o}*t+\frac{1}{2}*a*t^{2} \\x-x_{o}=(v_{f}-a*t)*t+\frac{1}{2} *a*t^{2}\\70=(15-a*t)*t+\frac{1}{2}*a*t^{2}\\70=15*t-a*t^{2} +\frac{1}{2}*a*t^{2} \\70=15*t-\frac{1}{2}*a*t^{2}\\70=15*(7)-\frac{1}{2} *a*(7)^{2}\\105-70=0.5*a*49\\35=24.5*a\\a=1.428[m/s^{2} ]](https://tex.z-dn.net/?f=x-x_%7Bo%7D%3Dv_%7Bo%7D%2At%2B%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E%7B2%7D%20%5C%5Cx-x_%7Bo%7D%3D%28v_%7Bf%7D-a%2At%29%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2At%5E%7B2%7D%5C%5C70%3D%2815-a%2At%29%2At%2B%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E%7B2%7D%5C%5C70%3D15%2At-a%2At%5E%7B2%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E%7B2%7D%20%5C%5C70%3D15%2At-%5Cfrac%7B1%7D%7B2%7D%2Aa%2At%5E%7B2%7D%5C%5C70%3D15%2A%287%29-%5Cfrac%7B1%7D%7B2%7D%20%2Aa%2A%287%29%5E%7B2%7D%5C%5C105-70%3D0.5%2Aa%2A49%5C%5C35%3D24.5%2Aa%5C%5Ca%3D1.428%5Bm%2Fs%5E%7B2%7D%20%5D)
Now replacing this value in the second equation, we can find the initial velocity.
![15=v_{o}+1.428*7\\v_{o}=5[m/s]](https://tex.z-dn.net/?f=15%3Dv_%7Bo%7D%2B1.428%2A7%5C%5Cv_%7Bo%7D%3D5%5Bm%2Fs%5D)
