Position displacement velocity acceleration are vectors and the rest are scalars
Answer:
frictional force = 0.52 N
Explanation:
diameter of turn table (D1) = 30 cm = 0.3 m
mass of turn table (M1) = 1.2 kg
diameter of shaft (D2) = 1.2 cm = 0.012 m
mass of shaft (M2) = 450 g = 0.45 kg
time (t) = 15 seconds
acceleration due to gravity (g) = 9.8 m/s^{2}
radius of turn table (R1) = 0.3 / 2 = 0.15 m
radius of shaft (R2) = 0.012 / 2 = 0.006 m
total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft
I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}
I = 0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}
I = 0.0135 + 0.0000081 = 0.0135081
ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s
α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}
torque = I x α
torque = 0.0135081 x (-0.23) = - 0.00311 N.m
torque = frictional force x R2
- 0.00311 = frictional force x 0.006
frictional force = 0.52 N
Answer:
v = -14 m/s
Explanation:
Given that,
Initial location of the ball, X₁ = 10 m
Final position of the ball, X₂ = -25 m
Time taken to travel is, t = 2.5 s
The average velocity of the ball is given by the formula,
V = X₂ - X₁ / t m/s
Substituting the values in the above equation,
V = -25 - 10 / 2.5
= -14 m/s
The negative sign in the velocity indicates that ball rolls in the opposite direction.
Hence, the average velocity of the ball is v = -14 m/s
Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
'G' = 9.0 m/s²
M = (0.08333 / 0.1755) (9.8) = 4.65 meters .
That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .
Answer:
1.4 * 10 ^-1 Ω
Explanation:
Hi,
For this question, we gotta use the formula
R = pL/A
p = The resistivity of your material at 20°C
L = length of the wire
A = cross-sectional area
The resistivity of tungsten is 5.60 * 10^-8 at 20°C
By plugging the values, we get:
R = (5.60 * 10^-8)(2.0)/(7.9*10^-7) = 1.4 * 10 ^-1 Ω
