Answer:
a. True
Explanation:
Metals are amazing, and they are all around us. You can probably easily identify them by their shiny surfaces and tinny sound when you tap them.
Generally, metals are mainly hard, shiny, and can be hammered into thin sheets. Also, another property of a metal is that, it is a good conductor of heat and even electricity because there valence (outer) electrons are delocalized. Some examples of chemical elements that are metals are iron, gold, sodium, silver, copper etc.
An accelerometer has a damping ratio of 0.5 and a natural frequency of 18,000 Hz. It is used to sense the relative displacement of a beam to which it is attached. (a)If an impact to the beam imparts a vibration at 4500 Hz, calculate the dynamic error and phase shift in the accelerometer output. (b)Calculateits resonance frequency.(c)What isthe maximumpossiblemagnitude ratio that the system can achieve
Answer:
A) i) Dynamic error ≈ 3.1%
ii) phase shift ≈ -12°
B) 79971.89 rad/s
Explanation:
Given data :
Damping ratio = 0.5
natural frequency = 18,000 Hz
a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz
i) Dynamic error
This can be calculated using magnitude ratio formula attached below is the solution
dynamic error ≈ 3.1%
ii) phase shift
This phase shift can be calculated using frequency dependent phase shift formula
phase shift ≈ -12°
B) Determine resonance frequency
Wr = 2[tex]\pi[/tex] ( 18000 [tex]\sqrt{0.5}[/tex] ) = 79971.89 rad/s
C) The maximum magnitude ratio that the system can achieve
A particle move in the xy plane so that its position vector r=bcosQi +bsinQj+ ctk, where b, Q and c are constants. show that the partial move with constant speed.
Answer:
The speed of this particle is constantly [tex]c[/tex].
Explanation:
Position vector of this particle at time [tex]t[/tex]:
[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].
Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:
[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].
Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].
Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:
[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].
The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :
[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].
Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)
2. Using a giant screw, a crew does 650 J of work to drill a hole into a rock.
The screw does 65 J of work. What is the efficiency of the screw? Show your
work. Hellpppp
Answer:
42,250
Explanation:
It goes inside=
Displacemt
It does work=
Work done
To find efficiency of jule we do=
Dicplacement × Work done
650 × 65
42,250
Please mark me as a brainlist
A soccer ball was kicked over the edge of a wall and traveled 35 m horizontally at a speed of 5.6m/s. Calculate the vertical height of the wall.
Answer:
Are you sure it was soccer ball? Or meine hearts
Explanation:
Given that Carbon-14 has a half-life of 5700 years, determine how long it would take for
this reduction to occur.
Answer:It will take about 3000 years
Explanation:
Derase
An electric heater Consumes 1.8 MJ When connected to a 250V supply for 30 minutes. Find the power rating of the heater and the current taken from the supply
Answer:
a. Power = 1000 Watts or 1 Kilowatts.
b. Current = 4 Amperes.
Explanation:
Given the following data;
Energy consumed = 1.8MJ = 1.8 × 10^6 = 1800000 Joules
Voltage = 250V
Time = 30 minutes to seconds = 30 * 60 = 1800 seconds
To find the power rating;
Power = energy/time
Substituting into the equation, we have;
Power = 1800000/1800
Power = 1000 Watts or 1 Kilowatts.
b. To find the current taken from the supply;
Power = current * voltage
1000 = current * 250
Current = 1000/250
Current = 4 Amperes.
An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object? Neglect air resistance.
how many pennies can 4 folds of a paper hold?
if a car travels 200 m to the east in 8.0 s what is the cars average velocity?
Answer:
25 m/s
Explanation:
200/8 = 25
Which is an example of kinetic energy?
A. The energy stored in
ethanol
B. A ball sitting at the top of a ramp
C. A compressed spring
D. A hockey puck sliding across ice
D. A hockey puck sliding across ice
is 250 000 miles from the earth to the moon" is a qualitative
Observation
TRUE
Or false
Answer:
True
Explanation:
Help plsssssssssss I write it 100 time no one answer
Answer:
1.93×10²⁸ s
Explanation:
From the question given above, the following data were obtained:
Number of electron (e) = 2×10²⁴
Current (I) = 10 A
Time (t) =?
Next, we shall determine the quantity of electricity flowing through pasing through the point. This can be obtained as follow:
1 e = 96500 C
Therefore,
2×10²⁴ e = 2×10²⁴ e × 96500 / 1 e
2×10²⁴ e = 1.93×10²⁹ C
Thus, 1.93×10²⁹ C of electricity is passing through the point.
Finally, we shall determine the time. This can be obtained as follow:
Current (I) = 10 A
Quantity of electricity = 1.93×10²⁹ C
Time (t) =?
Q = it
1.93×10²⁹ = 10 × t
Divide both side by 10
t = 1.93×10²⁹ / 10
t = 1.93×10²⁸ s
Thus, it took 1.93×10²⁸ s for 2×10²⁴ electrons to pass through the point