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2 years ago

9

a hiker walks 200 m west and then walks 100 m north in what direction is her resulting displacement draw to show your answer

1 answer:

Sliva [168]2 years ago

5 0

Answer:

The direction of the displacement is in North-West.

Explanation:

Resultant displacement D is

$=\sqrt{(200)^{2} + (100)^{2} } \\=223.60m$

Here the direction is

$\Theta =tan^{^{-1}}\left ( \frac{100}{200} \right )\\\Theta =26.6^{o}$

Then the direction is $26.6^{o}$ North-west.

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A 12 oz can of soda is left in a car on a hot day. In the morning the soda temperature was 60oF with a gauge pressure of 40 psi.

Neko [114]

In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:

P/T = Constant

Then
$\frac{ P_{1} }{ T_{1} } = \frac{ P_{2} }{ T_{2} }$

Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)

Substituting;
$P_{2} = \frac{ P_{1} T_{2} }{ T_{1} } = \frac{40*305}{289} =42.21 psi$

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3 years ago

A 2.5m long steel piano wire has a diameter of 0.5cm how great is the tention in the wire if it stretches by 0.45cm when tighten

UkoKoshka [18]

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Explanation:

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3 years ago

A 6.00-μf parallel-plate capacitor has charges of 40.0 μc on its plates. how much potential energy is stored in this capacitor?

dedylja [7]

Thew energy stored in a capacitor of capacitance $C$ and voltage between the plates $V$ is

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3 years ago

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

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3 years ago

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Andrei [34K]

Answer:

Im pretty sure 1 is gravity, 2 is force

Explanation:

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