Answer:
is his final displacement from the point A after 60 seconds.
Explanation:
Given:
Cyclist is moving away from A.
- velocity of cyclist,
- displacement of the cyclist from point A at the time of observation,
- time after which the next observation is to be recorded,
Now as the cyclist is moving away from point A his change in displacement after the mentioned time:
<u>Now the the final displacement from point A after the mentioned time:</u>
The far side of the Moon is the hemisphere of the Moon that always faces away from Earth. The far side's terrain is rugged with a multitude of impact craters and relatively few flat lunar maria compared to the near side. It has one of the largest craters in the Solar System, the South Pole–Aitken basin.
Answer:
a) 
.
b) 
.
Explanation:
a) What is the speed of the rock just before it hits the street?
From Kinematics we have 
.
If we take the top of the roof as the position 
, then
and we have (also, 
) :
⇒ 
⇒ .
b) How much time elapses from when the rock is thrown until it hits the street?
From Kinematics we have 
When the rock touches the ground:
With a minus sign to indicate the vector velocity points down.
(remember that 
)
The constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
<h3> Constant angular acceleration</h3>
Apply the following kinematic equation;
ωf² = ωi² - 2αθ
where;
- ωf is the final angular velocity when the centrifuge stops = 0
- ωi is the initial angular velocity
- θ is angular displacement
- α is angular acceleration
ωi = 3400 rev/min x 2π rad/rev x 1 min/60s = 356.05 rad/s
θ = 52 rev x 2π rad/rev = 326.7 rad
0 = ωi² - 2αθ
α = ωi²/2θ
α = ( 356.05²) / (2 x 326.7)
α = 194.02 rad/s²
Thus, the constant angular acceleration (in rad/s2) of the centrifuge is 194.02 rad/s².
Learn more about angular acceleration here: brainly.com/question/25129606
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Answer:
0.8 N
Explanation:
From coulomb's law,
Formula:
F = kqq'/r²........................ Equation 1
Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.
Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.
Constant: k = 9×10⁹ Nm²/C²
Substitute these values into equation 1
F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²
F = 1800×10⁻³/2.25
F = 1.8/2.25
F = 0.8 N
