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Citrus2011 [14]
3 years ago
5

When an object is dropped from the top of a tall cliff, your teacher tells you to use -10 m/s2 for the approximate acceleration

due to gravity. What sign should the displacement of the object have and why?
A) The displacement should be positive, because it only moves in one direction.
B) The displacement should be positive, because this choice would make calculations easier
C) The displacement should be negative, because the negative represent a downward direction.
D) The displacement should be positive, because you can never have less than zero displacement.

The truck is described as travelling at 30mph to the west. The quantity described here is a(n)
A) speed
B) velocity
C) acceleration
D) kinetic energy
Physics
2 answers:
N76 [4]3 years ago
6 0

1- C

2-B

Those are your answers hope this helps :)


Romashka-Z-Leto [24]3 years ago
3 0

1. Answer: Option C

Displacement is a vector quantity. It has both magnitude as well as direction. Hence, it can be positive,  zero or negative. In the given question, downward direction is taken as negative. This we have come to know from the downward acceleration -10m/s^2. The negative sign indicates that downward direction is considered negative.

Now, since the object is dropped from the top. The displacement should be negative, because the negative represent a downward direction.

2. Answer: velocity

The truck is travelling at 30 mph to the west. From the unit mph (miles per hour), we know its a magnitude of velocity. West indicates the direction. Hence, the given quantity is a vector -velocity.

It is not speed because speed only indicates the magnitude and not the direction. speed is a scalar quantity.

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Which of the following is represented by the letter A in the diagram below?
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We have that the letter A in the diagram below given as

Amplitude

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<h3>Amplitude</h3>

Question Parameters:

Amplitude

Crest

Trough

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Amplitude is denoted with the letter A

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2 years ago
If you have two uncertainties, and they are from two different sources and contribute to the uncertainty of a measurement, what
Darya [45]

The propagation errors we can find the uncertainty of a given magnitude is the sum of the uncertainties of each magnitude.

                           Δm = ∑  | \frac{dm}{dx_i} | \ \Delta x_i

Physical quantities are precise values ​​of a variable, but all measurements have an uncertainty, in the case of direct measurements the uncertainty is equal to the precision of the given instrument.

When you have derived variables, that is, when measurements are made with different instruments, each with a different uncertainty, the way to find the uncertainty or error is used the propagation errors to use the variation of each parameter, keeping the others constant and taking the worst of the  cases, all the errors add up.

If m is the calculated quantity, x_i the measured values ​​and Δx_i the uncertainty of each value, the total uncertainty is

                      Δm = ∑  | \frac{dm}{dx_i } | \ \Delta x_i    | dm / dx_i | Dx_i

               

for instance:

If the magnitude is  a average of two magnitudes measured each with a different error

                     m = \frac{m_1+m_2}{2}

                     Δm = | \frac{dm}{dx_1} |  Δx₁ + | \frac{dm}{dx_2} | Δx₂

                     \frac{dm}{dx_1} = ½

                     \frac{dm}{dx_2} = ½

                     Δm = \frac{1}{2} Δx₁ + ½ Δx₂

                     Δm = Δx₁ + Δx₂

In conclusion, using the propagation errors we can find the uncertainty of a given quantity is the sum of the uncertainties of each measured quantity.

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