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14 hours ago

Ingrid is moving a box from the ground into the back of a truck. She uses 20 N of force to move the box 5 meters. If she uses an

inclined plane to reduce her force to 10 N which of the following statements will be true?
A. She will decrease the amount of work done.
B. She will make her work efficiency 100 percent.
C. She will increase the distance the box moves.
D. All of the above.

Physics

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Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan

Sav [3153]

Based on my analysis, the image aligns well with the inquiry. From the provided graph, we can derive the equation of the line through the two-point formula. Knowing the temperature at 601 K allows us to calculate the corresponding altitude.

y - y1 = [(y2 - y1)/(x2 - x1)](x - x1)
y - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4)

By substituting y = 601, we can solve for x:
601 - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4<span>)
</span>x = 83

Hence, the instruments on the probe will fail at an altitude of 83 kilometers.

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2 months ago

Richard needs to fly from san diego to halifax, nova scotia and back in order to give an important talk about mathematics. on th

Ostrovityanka [3204]

As the plane heads toward Halifax, the wind speed supports the flight path

resulting in an overall improved speed

Conversely, during the return trip, the wind will resist the plane's motion, decreasing the net speed

The total journey lasts 13 hours

of which 2 hours was dedicated to the mathematics discussion

Consequently, the total flight time is 13 - 2 = 11 hours

Now we apply the formula to calculate the time for traveling to Halifax

$t_1 = \frac{d}{v + 50}$

Time needed to return

$t_2 = \frac{d}{v - 50}$

Let’s look at the total time

$T = t_1 + t_2$

$11 = \frac{d}{v - 50} + \frac{d}{v + 50}$

Here d = 3000 miles

$11 = \frac{3000}{v - 50} + \frac{3000}{v + 50}$

$3.67 * 10^{-3} = \frac{2v}{v^2 - 2500}$

$v^2 - 2500 = 545.45v$

By solving the derived quadratic equation

$v = 550 mph$

the plane's speed calculates to 550 mph

3 0

2 months ago

Read 2 more answers

What will be the final temperature if a 4.00 g silver ring at 41.0◦C if it gives off 18.0 J of heat to the surroundings? The spe

kicyunya [3294]

Final temperature to determine: Given the following details, the calculations proceed as follows: Mass of the silver ring is m = 4 g, initial temperature is presented, and the heat released is Q = -18 J (indicating heat loss). The specific heat of silver is considered next to find the final temperature.

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2 months ago

On its own, a certain tow-truck has a maximum acceleration of 3.0 m/s2. what would be the maximum acceleration when this truck w

inna [3103]

Let us denote $a_1=3 m/s^2$ as the highest acceleration of the truck by itself. The force generated by the engine to propel the truck in this scenario is
$F= ma_1$
where m is the mass of the truck alone.

Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is
$F=3m a_2$
where a2 represents the new acceleration.
Since the engine remains unchanged, the force generated stays the same, allowing us to set the force equations equal to each other:
$m a_1 = 3 m a_2$
and solving for a2 gives us:
$a_2 = \frac{m a_1}{3 m}= \frac{a_1}{3}= \frac{3 m/s^2}{3}=1 m/s^2$

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3 months ago

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A point charge with charge q1 is held stationary at the origin. A second point charge with charge q2 moves from the point (x1, 0

kicyunya [3294]

Answer:

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Explanation:

The position of the charge q₁ is established at (0,0)

Meanwhile, the charge q₂ is located at (x₁,0)

Thus, the electric potential energy between these two charges is determined by:

$U_1=k\dfrac{q_1q_2}{x_1}$

Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:

$U_2=k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}}$

According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:

$W=-\Delta U$

$W=-(U_2-U_1)$

$W=(U_1-U_2)$

$W=(k\dfrac{q_1q_2}{x_1}-k\dfrac{q_1q_2}{\sqrt{x_2^2+y_2^2}})$

$W=kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$

Consequently, the work done by the electrostatic force on the moving charge is $kq_1q_2(\dfrac{1}{x_1}-\dfrac{1}{\sqrt{x_2^2+y_2^2}})$ . Therefore, this concludes the solution.

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2 months ago