B strength training I think that’s the answer
Answer:
W = 0.49 N
τ = 0.4851 Nm
Force
Explanation:
The weight force can be found as:
W = mg
W = (0.05 kg)(9.8 m/s²)
<u>W = 0.49 N</u>
The torque about the pivot can be found as:
τ = W*d
where,
τ = torque
d = distance between weight and pivot = 99 cm = 0.99 m
Therefore,
τ = (0.49 N)(0.99 m)
<u>τ = 0.4851 Nm</u>
The pivot exerts a <u>FORCE </u>on the meter stick because the pivot applies force normally over the stick and has a zero distance from stick.
Thank you for posting your question here at brainly. Below is the solution. I hope the answer will help.
<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9
</span>
S = 2 + 6.8 + 2.45 = 11.25
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>
consider the motion along the horizontal direction :
v₀ = initial velocity in horizontal direction as the ball rolls off the table = 3.0 m/s
X = horizontal displacement of the ball = 2.0 m
a = acceleration along the horizontal direction = 0 m/s²
t = time taken to land = ?
using the kinematics equation
X = v₀ t + (0.5) a t²
2.0 = 3.0 t + (0.5) (0) t²
t = 2/3
consider the motion of the ball along the vertical direction
v₀ = initial velocity in vertical direction as the ball rolls off the table = 0 m/s
Y = vertical displacement of the ball = height of the table = h
a = acceleration along the vertical direction = 9.8 m/s²
t = time taken to land = 2/3
using the kinematics equation
Y = v₀ t + (0.5) a t²
h = 0 t + (0.5) (9.8) (2/3)²
h = 2.2 m
C 2.2 m
