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mote1985 [20]
3 years ago
8

An object is represented by the dot on a motion map. What is the best description for the motion of this object? The object is m

oving with a constant speed. The object is moving with a constant acceleration. The object is standing still. The object is falling.
Physics
2 answers:
ankoles [38]3 years ago
5 0

The object is standing still.

alexira [117]3 years ago
5 0

Answer:

C. The object is standing still

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PLEASE SOMEONE ANSWER!!! thank you so so so much!!!!
Oduvanchick [21]

Answer:

As much I know the gravity on moon is 1.62m/s२.

7 0
4 years ago
Tim and Rick both can run at speed Vr and walk at speed Vw, with Vr > Vw.
miss Akunina [59]

Answer:

Δt =  \frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw}

Explanation:

Hi there!

Using the equation of speed for the whole trip, we can obtain the time each one needed to cover the distance D.

The speed (v) is calculated by dividing the traveled distance (d) over the time needed to cover that distance (t):

v = d/t

Rick traveled half of the distance at Vr and the other half at Vw. Then, when v = Vr, the distance traveled was D/2 and the time is unknown, Δt1:

Vr = D/ (2 · Δt1)

For the other half of the trip the expression of velocity will be:

Vw = D/(2 · Δt2)

The total time traveled is the sum of both Δt:

Δt(total) = Δt1 + Δt2

Then, solving the first equation for Δt1:

Vr = D/ (2 · Δt1)

Δt1 = D/(2 · Vr)

In the same way for the second equation:

Δt2 = D/(2 · Vw)

Δt + Δt2 = D/(2 · Vr) + D/(2 · Vw)

Δt(total) = D/2 · (1/Vr + 1/Vw)

The time needed by Rick to complete the trip was:

Δt(total) = D/2 · (1/Vr + 1/Vw)

Now let´s calculate the time it took Tim to do the trip:

Tim walks half of the time, then his speed could be expressed as follows:

Vw = 2d1/Δt  Where d1 is the traveled distance.

Solving for d1:

Vw · Δt/2 = d1

He then ran half of the time:

Vr = 2d2/Δt

Solving for d2:

Vr · Δt/2 = d2

Since d1 + d2 = D, then:

Vw · Δt/2 +  Vr · Δt/2 = D

Solving for Δt:

Δt (Vw/2 + Vr/2) = D

Δt = D / (Vw/2 + Vr/2)

Δt = D/ ((Vw + Vr)/2)

Δt = 2D / (Vw + Vr)

The time needed by Tim to complete the trip was:

Δt = 2D / (Vw + Vr)

Let´s find the diference between the time done by Tim and the one done by Rick:

Δt(tim) - Δt(rick)

2D / (Vw + Vr) - (D/2 · (1/Vr + 1/Vw))

\frac{2D}{Vw+Vr} - \frac{D}{2Vr} - \frac{D}{2Vw} = Δt

Let´s check the result. If Vr = Vw:

Δt = 2D/2Vr - D/2Vr - D/2Vr

Δt = D/Vr - D/Vr = 0

This makes sense because if both move with the same velocity all the time both will do the trip in the same time.

8 0
4 years ago
Qué representa el lanzamiento vertical ascendente
rusak2 [61]

Answer:

Uhhhhh? can u explains? lol

6 0
3 years ago
WHOEVER ANSWERS FIRST GETS BRAINLEST
JulsSmile [24]

<u>Communications:</u> Laura develops programs that flag suspicious network activity.

<h3>What is communication?</h3>

Communication involves the transfer of information from one person (sender) to another (recipient), especially through the use of semiotics, symbols, signs and network devices.

<h3>How to match the scenarios?</h3>
  • <u>Communications:</u> Laura develops programs that flag suspicious network activity.
  • <u>Healthcare:</u> Greg calibrates electronic scales.
  • <u>Manufacturing:</u> Tim ensures the output of a process remains in a defined range.

Read more on communication here: brainly.com/question/26152499

#SPJ1

8 0
2 years ago
100 kw of power is delivered to the other side of a city by a pair of power lines with the voltage difference of 13014.1 v.
FinnZ [79.3K]
A) The power delivered to the lines is
P_{in}= 100 kW=1 \cdot 10^5 W
And the voltage at which the lines work is
V=13014.1 V
Since the power delivered is the product between the voltage and the current:
P=VI
We can find the current flowing in the lines:
I= \frac{P}{V}= \frac{1 \cdot 10^5 W}{13014.1 V}=7.68 A

b) The voltage change along each line can be found by using Ohm's law:
\Delta V = IR = (7.68 A)(10 \Omega)=76.8 V

c) The power wasted as heat along each line is given by:
P_d = I^2 R = (7.68 A)^2 (10 \Omega) = 590 W
And since we have 2 lines, the total power wasted as heat in both lines is
P_d = 2 \cdot 590 W=1180 W
6 0
3 years ago
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