Before we go through the questions, we need to calculate and determine some values first.
r = 11.5 m
<span>m = 280 kg </span>
<span>Centripetal force = m x v^2/r = 280 x (17.1^2/11.5) = 7119.55 N
</span>
1) What is the magnitude of the normal force on the care when it is at the bottom of the circle.
<span>Centripetal force + mg = 7119.55 + (280 x 9.8) = 9863.55 N </span>
<span>2) What is the magnitude of the normal force on the car when it is at the side of the circle. </span>
<span>Centripetal force = 7119.55 N </span>
<span>3) What is the magnitude of the normal force on the car when it is at the top of the circle. </span>
<span>Centripetal force - mg = 7119.55 - (280 x 9.8) = 4375.55 N </span>
<span>4) What is the minimum speed of the car so that it stays in contact with the track at the top of the loop. </span>
√<span>(gr) </span>
√<span>(9.8 x 11.5) = 10.62 m/s</span>
Answer:
C. 100
D.3
E. 33.3
Explanation:
C. Mechanical Advantage=Load / Effort
= 200N
--------
100N
Therefore,. = 100
D. I. Velocity Ratio= distance moved by the effort / distance moved by load
= 30cm/10cm
= 3
II. Efficiency= M.A / V.R
= 100/3
= 33.33
C isotopes is the correct answer. Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number.All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.
Hope this helps! :)
Answer:
F = 85696.5 N = 85.69 KN
Explanation:
In this scenario, we apply Newton's Second Law:
where,
F = Upthrust = ?
m = mass of space craft = 5000 kg
g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)
g = 15.0093 m/s²
a = acceleration required = 2.13 m/s²
Therefore,
<u>F = 85696.5 N = 85.69 KN</u>
Answer:
ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN ZAMN
Explanation:
