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3 years ago

How do constellations differ from other groupings of stars such as galaxies?

1 answer:

3 0

I think a constellation is an area of sky that contains up a picture, object, animal, ect. - a galaxy is a grouping of several million to billion stars, gas and dust, that is gravitationally bound.

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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

SOVA2 [1]

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

<u>a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz</u>

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°

<u>B) Determine resonance frequency </u>

Wr = 2 $\pi$ ( 18000 $\sqrt{0.5}$ ) = 79971.89 rad/s

C) The maximum magnitude ratio that the system can achieve

3 0

3 years ago

a school bus has stopped to allow children to get off the bus which graph shows the motion of the bus?

Usimov [2.4K]

Answer: what is it in

Explanation:

4 0

3 years ago

At a distance of 10 km from a radio transmitter, the amplitude of the electric field is 0. 20 volts/meter. What is the total pow

Nady [450]

a. 10 kW b. 67 kW c. 140 kW d. 245 kW e. 21 kW

8 0

3 years ago

A ball is thrown vertically downward from the top of a 37.4-m-tall building. The ball passes the top of a window that is 15.4 m

Effectus [21]

Answer:

v= 20.8 m/s

Explanation:

Assuming no other forces acting on the ball, from the instant that is thrown vertically downward, it's only accelerated by gravity, in this same direction, with a constant value of -9.8 m/s2 (assuming the ground level as the zero reference level and the upward direction as positive).
In order to find the final speed 2.00 s after being thrown, we can apply the definition of acceleration, rearranging terms, as follows:

$v_{f} = v_{o} + a*t = v_{o} + g*t (1)$

We have the value of t, but since the ball was thrown, this means that it had an initial non-zero velocity v₀.
Due to we know the value of the vertical displacement also, we can use the following kinematic equation in order to find the initial velocity v₀:

$\Delta y = v_{o} *t + \frac{1}{2} * a* t^{2} (2)$

where Δy = yf - y₀ = 15.4 m - 37.4 m = -22 m (3)

Replacing by the values of Δy, a and t, we can solve for v₀ as follows:

$v_{o} = \frac{(\Delta y- \frac{1}{2} *a*t^{2})}{t} = \frac{-22m+19.6m}{2.00s} = -1.2 m/s (4)$

Replacing (4) , and the values of g and t in (1) we can find the value that we are looking for, vf:

$v_{f} = v_{o} + g*t = -1.2 m/s - (9.8m/s2*2.00s) = -20.8 m/s (5)$

Therefore, the speed of the ball (the magnitude of the velocity) as it passes the top of the window is 20.8 m/s.

4 0

3 years ago

Explain why it is not advisable to use small values of i in performing triangular glass prism experiment

kifflom [539]

Is this about using triangular prisms in spectrometer experiments ? See enclosed

7 0

4 years ago