Instantaneous impulse and continuous force: these are the foundations of Newton laws. In short explanation; concept of force and mass.
<span>You are given a QL = -26 μC charge that is placed on the x-axis at x = - 0.2 m and a QR = 26 μC charge that is placed at x = +0.2 m. The answers are:
The x-component of the electric field at x = 0 m and y = 0.2 m is 3.
The y-component of the electric field at x = 0 m and y = 0.2 m is 2.
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Answer:
Explanation:
cSep 20, 2010
well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.
Anonymous
Sep 20, 2010
First you need to solve for time by using
d=(1/2)(a)(t^2)+(vi)t
1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s
t=.45 sec
Then you find the horizontal distance traveled by using
v=d/t
1.3m/s=d/.54sec
d=.585m
Then you need to find the time of player B by using
d=(1/2)(a)(t^2)+(vi)t
1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0
t=.61 sec
Finally to find player Bs initial horizontal velocity you use the horizontal equation
v=d/t
v=.585m/.61 sec
so v=.959m/s
Answer:
x = 6.94 m
Explanation:
For this exercise we can find the speed at the bottom of the ramp using energy conservation
Starting point. Higher
Em₀ = K + U = ½ m v₀² + m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
½ m v₀² + m g h = ½ m v²
v² = v₀² + 2 g h
Let's calculate
v = √(1.23² + 2 9.8 1.69)
v = 5.89 m / s
In the horizontal part we can use the relationship between work and the variation of kinetic energy
W = ΔK
-fr x = 0- ½ m v²
Newton's second law
N- W = 0
The equation for the friction is
fr = μ N
fr = μ m g
We replace
μ m g x = ½ m v²
x = v² / 2μ g
Let's calculate
x = 5.89² / (2 0.255 9.8)
x = 6.94 m