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Lubov Fominskaja [6]
2 years ago
6

What does the object below model?

Physics
1 answer:
harina [27]2 years ago
3 0

Answer:  c

Explanation: it is c because i used my brain to answer it

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A metal ion (X) with a charge of 4+ is attracted to a nonmetal ion (Z) with a
Keith_Richards [23]

Answer:

A) X3Z4

Explanation:

8 0
2 years ago
How much pressure is on the bottom of a pot that holds 20N of soup? the surface area of the pot is 0.05m2
fomenos

Answer:

400

Explanation:

Formula used in solution:

P = \frac{F}{A}

P - pressure,\: F - force,\:  A - area

The given information:

F = 20N

A = 0.05m^{2}

Solution

P = \frac{F}{A} = \frac{20}{0.05} = 20 * 20 = 400

Answer:

Pressure = 400 \: Pascals

8 0
3 years ago
An image formed on a screen is always​
Annette [7]

Answer:

diminished and erect( upright)

Explanation:

6 0
2 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
3 years ago
Lead-202 has a half-life of 53,000 years. How long will it take for 15/16 of a sample of lead-202 to decay?
ohaa [14]

Answer:

C. 212,000 years

Explanation:

believe me it's correct.....and you're welcome :)

4 0
3 years ago
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