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2 years ago

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A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =

9.00î m/s.(a) find the vector position of the particle at any time t (where t is measured in seconds).

1 answer:

Mkey [24]2 years ago

3 0

The position S(t) at any time of the particle can be written as
$S(t)=S_0 + vt + \frac{1}{2}at^2$
where $S_0$ is the initial position (in our case, the particle is initially located at the origin, so S0=0i+0j, v is the velocity, a the acceleration and t the time.

Using v(t)=9.00 i m/s and a(t)=2.00 j m/s^2, we can rewrite S(t), the vector position of the particle at time t:
$S(t)=9t \mbox{\bf{i}} m/s + \frac{1}{2} 2 t^2 \mbox{\bf{j}} m/s^2$

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

evablogger [386]

Answer:

$\text{heat loss} = 24864.05 \ W/m^2$

Explanation:

If

$T_1$ , $T_2$ are temperatures of gasses and liquid in Kelvins,

$t_1$ and $t_2$ are thicknesses of gas layer and steel slab in meters,
$h_1$ , $h_2$ are convection coefficients gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ is the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ are thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

using known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Using the rate equation :

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Similarly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature distribution is shown in the attached image

3 0

2 years ago

A dolphin swims due east for 1.90 km, then swims 7.20 km in the direction south of west. What are the magnitude and direction of

kykrilka [37]

Answer:

magnitude = 7.446 km, direction = 75.22° north of east

Explanation:

From the questions,

To get the the magnitude of the resultant vector we use Pythagoras theorem

a² = b²+c²

From the diagram,

y² = 1.9²+7.2²

y² = 55.45

y = √(55.45)

y = 7.446 km.

The direction of the dolphin is given as,

θ = tan⁻¹(7.2/1.9)

θ = tan⁻¹(3.7895)

θ = 75.22° north of east

Hence the magnitude of the resultant vector = 7.446 km, and it direction is 75.22° north of east

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2 years ago

Step 8: Observe How Changes in the Speed of the Bottle Affect Beanbag Height

lina2011 [118]

Answer:

When the speed of the bottle is 2 m/s, the average maximum height of the beanbag is <u>0.10</u> m.

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When the speed of the bottle is 4 m/s, the average maximum height of the beanbag is <u>0.87</u> m.

When the speed of the bottle is 5 m/s, the average maximum height of the beanbag is <u>1.25</u> m.

When the speed of the bottle is 6 m/s, the average maximum height of the beanbag is <u>1.86</u> m.

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6 0

2 years ago

Read 2 more answers

In Part 6.2.2, you will determine the wavelength of the laser by shining the laser beam on a "diffraction grating", a set of reg

harkovskaia [24]

Answer:

λ = 2042 nm

Explanation:

given data

screen distance d = 11 m

spot s = 4.5 cm = 4.5 × $10^{-2}$ m

separation L = 0.5 mm = 0.5 × $10^{-3}$ m

to find out

what is λ

solution

we will find first angle between first max and central bright

that is tan θ = s/d

tan θ = 4.5 × $10^{-2}$ / 11

θ = 0.234

and we know diffraction grating for max

L sinθ = mλ

here we know m = 1 so put all value and find λ

L sinθ = mλ

0.5 × $10^{-3}$ sin(0.234) = 1 λ

λ = 2042.02 × $10^{-9}$ m

λ = 2042 nm

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2 years ago

An electric eel (Electrophorus electricus) can produce a shock of up to 600 V and a current of 1 A for a duration of 2 ms, which

Irina-Kira [14]

Answer:

$2\times 10^{-3}\ C$

6000

1.2 J

$3.33\times 10^{-6}\ F$

Explanation:

I = Current = 1 A

t = Time = 2 ms

n = Number of electrocyte

V = Voltage = 100 mV

Charge is given by

$Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C$

The charge flowing through the electrocytes in that amount of time is $2\times 10^{-3}\ C$

The maximum potential is given by

$V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000$

The number of electrolytes is 6000

Energy is given by

$E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J$

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

$C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F$

The equivalent capacitance of all the electrocyte cells in the electric eel is $3.33\times 10^{-6}\ F$

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2 years ago