Answer:
The answer is A. 10
Explanation:
<em>Given </em>
<em>f(</em><em>x</em><em>)</em><em> </em><em>=</em><em> </em><em>3x </em><em>-</em><em> </em><em>1</em><em>1</em>
<em>So, </em><em> </em>
<em>f(</em><em>7</em><em>)</em><em> </em><em>=</em><em> </em><em>3</em><em> </em><em>*</em><em> </em><em>7</em><em> </em><em>-</em><em> </em><em>1</em><em>1</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em>1</em><em> </em><em>-</em><em> </em><em>1</em><em>1</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>1</em><em>0</em>
(a) The velocity ratio of the screw is 1570.8.
(b) The mechanical advantage of the screw is 785.39.
<h3>
Velocity ratio of the screw</h3>
The velocity ratio of the screw is calculated as follows;
V.R = 2πr/P
where;
- P is the pitch = 1/10 cm = 0.1 cm = 0.001 m
- r is radius = 25 cm = 0.25 m
V.R = (2π x 0.25)/(0.001)
V.R = 1570.8
<h3>Mechanical advantage of the screw</h3>
E = MA/VR x 100%
0.5 = MA/1570.8
MA = 785.39
Learn more about mechanical advantage here: brainly.com/question/18345299
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Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
(3) The period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4) he gravitational force between the Sun and Neptune is 6.75 x 10²⁰ N
Explanation:
(3) The period of a satellite is given as;

where;
T is the period of the satellite
M is mass of Earth
r is the radius of the orbit
Thus, the period of the satellite is independent of its mass, an increase in the mass of the satellite will not affect its period around the Earth.
(4)
Given;
mass of the ball, m₁ = 1.99 x 10⁴⁰ kg
mass of Neptune, m₂ = 1.03 x 10²⁶ kg
mass of Sun, m₃ = 1.99 x 10³⁰ kg
distance between the Sun and Neptune, r = 4.5 x 10¹² m
The gravitational force between the Sun and Neptune is calculated as;

answer:
6 ohms
Explanation:
if these two resistors are connected in series, the total resistance is the sum: 2+4 = 6 (ohms)