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2 years ago

Car moves 5km East,3km south,2km West, 1km north find displacement(please show me the way)

Physics

2 answers:

ddd [48]2 years ago

4 0

Answer:

3.605551275463989

Explanation:

solve using Pythagorean theorem

Helen [10]2 years ago

3 0

Answer:

heyy babe

want to have fun?

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In an electron cloud, an electron farther east away from the nucleus has?

vladimir2022 [97]

An electron that is far away from the nucleus have higher energy than an electron near the nucleus. Nucleus are positively charged and those electrons near it get attracted; those electrons gain kinetic energy hence reducing their internal energy. The electrons far from nucleus have low kinetic energy hence more internal energy.

8 0

2 years ago

If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

Damm [24]

Answer:

The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

Explanation:

As data is incomplete here, so by seeing the complete question from the search the data is

vx_0=1.1 x 10^6

ax=0 As acceleration is zero in the horizontal axis so

Equation of motion in horizontal direction is given as

$s_x=v_x_0 t$

$t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s$

Now for the vertical distance

vy_o=0

than the equation of motion becomes

$s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2$

Now using this acceleration the value of electric field is calculated as

$E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\$

Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation

$E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C$

So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.

5 0

3 years ago

If you place 1 C of positive charge on Earth and 1 C of negative charge on the moon, 384,500 km away, how much force would the p

ivolga24 [154]

Answer:

6.1 x 10^-8 newtons

Explanation:

F = 8.98 *109 *1*1/3845000002

4 0

1 year ago

A 62 kg bungee jumper jumps from a bridge. She is tied to a bungee cord whose unstretched length is 12 m. She falls a total of 3

Andrew [12]

Answer:

k = 104.46 N/m

Explanation:

Here we can use energy conservation

so we will have

initial gravitational potential energy = final total spring potential energy

as we know that she falls a total distance of 31 m

while the unstretched length of the string is 12 m

so the extension in the string is given as

$x = L - L_o$

$x = 31 - 12 = 19 m$

so we have

$mgH = \frac{1}{2}kx^2$

$62(9.81)(31) = \frac{1}{2}k (19^2)$

$k = 104.46 N/m$

5 0

2 years ago

A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over

kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air $\nu\ of air =19.31*10^{-6} m2/s$

$\rho = 1.048 kg/m3$

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

$Re =\frac{VD}{\nu}$

$= \frac{ 15*0.025}{19.31*10^{-6}}$

= 1.941 *10^4

drag coffecient is given as

$C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}$

solving for f_D

$f_D = C_D A_f*\frac{\rho v^2}{2}$

$=C_D D*\frac{\rho v^2}{2}$

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

$f_D = 1.1*0.025 *\frac{1.048*15^2}{2}$

f_D = =3.24 N/m

4 0

2 years ago

Car moves 5km East,3km south,2km West, 1km north find displacement(please show me the way)​

Car moves 5km East,3km south,2km West, 1km north find displacement(please show me the way)