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3 years ago

What net force would be needed to accelerate the box at 2.3 m/s^2

2 answers:

6 0

The pertinent equation here is F=ma. You haven't shared the mass of the box, so I will use M to represent that mass.

Then F = M(<span>2.3 m/s^2) (answer)</span>

Hoochie [10]3 years ago

4 0

The required force, in newtons, is

(2.3) x (the mass of the box, in kilograms) .

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elixir [45]

Is there any types of answer to get an idea

8 0

3 years ago

Read 2 more answers

Location A receives

Vlad1618 [11]

Location A receives more rainfall than Location B due to the rain shadow effect.

<u>Explanation</u>:

Rain shadow effect is caused due to the presence of mountains.
A rain shadow area is an area of land that has been forced to become dry, devoid of any vegetation growth due to the blockage of precipitation by mountains. These rain shadow areas will have a dry climate.
The other side of the mountain would receive plenty of precipitation and therefore would be flourished with plant growth. These areas will have a cool and wet climate.
In this case, Location A is on the other side of the mountain and so receives more rainfall or precipitation. Meanwhile, Location B is on the rain shadow region and so receives less rainfall.

7 0

3 years ago

find the potential energy of an aircraft weighing 10000 bs at 5000 ft true altitude and 125 kts true air speed

Cloud [144]

Answer:

$U=5*10^7ft.Ib$

Explanation:

From the question we are told that

Weight $W= 10000bs$

Altitude $H=5000ft$

Speed $V=125kts\\1kts=0.514m/s\\V=125*0.514=>64.25m/s$

Generally the equation for Potential energy ids mathematically given as

$Potential\ Energy\ U=mgh$

$U=Wh$

$U=10000*5000$

$U=5*10^7ft.Ib$

6 0

3 years ago

What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo

Phantasy [73]

Answer:

The magnetic field will be $\large{\dfrac{1.4 \times 10^{-4}}{d}} T$ , '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ( $\large{\overrightarrow{B}}$ ) at a distance ' $r$ ' due to a current carrying conductor carrying current ' $I$ ' is given by

$\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}$

where ' $\overrightarrow{dl}$ ' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current ' $I$ ', at a distance ' $d$ ' is given by

$\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}$

According to the figure if ' $I_{t}$ ' be the current carried by the top wire, ' $I_{b}$ ' be the current carried by the bottom wire and ' $2d$ ' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the $\bigotimes$ symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by $\bigodot$ symbol.