How many electrons point in 2 second in a wire carrying a current of 20mA ?

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel

Sonja [21]

consider east-west direction along X-axis and north-south direction along Y-axis

$V_{ra}$ = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

$V_{ag}$ = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

$V_{rg}$ = velocity of migrating robin relative to ground = ?

using the equation

$V_{rg}$ = $V_{ra}$ + $V_{ag}$

$V_{rg}$ = 12 j + 6.3 i

$V_{rg}$ = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

4 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago

A delivery truck travels 2.8 km North, 1.0 km East, and 1.6 km South. The final displacement from the origin is ___km to the ___

34kurt

Answer:

The final displacement from the origin is 1.6 km to the NE

Explanation:

The directions in which the delivery truck travels are;

1) 2.8 km North = 2.8· $\hat j$ , in vector form

2) 1.0 km East = 1.0· $\hat i$ , in vector form

3) 1.6 km South = -1.6· $\hat j$ , in vector form

Therefore, to find the final displacement, Δx, of the delivery truck, we add the individual displacements as follows;

Final displacement, Δd = 2.8· $\hat j$ + 1.0· $\hat i$ +(-1.6· $\hat j$ ) = 1.2· $\hat j$ + 1.0· $\hat i$

Final displacement, = 1.0· $\hat i$ + 1.2· $\hat j$

Where;

Δx = The displacement in the x-direction = 1.0· $\hat i$

Δy = The displacement in the y-direction = 1.2· $\hat j$

The magnitude of the resultant displacement vector is given as follows

$\left | d \right |$ = √((Δx)² + (Δy)²) = √(1² + 1.2²) ≈ 1.6 (To the nearest tenth)

The magnitude of the resultant displacement vector ≈ 1.6 km

The direction of the resultant vector is positive for both the east and north direction, therefore, the direction of the resultant vector = NE

Therefore, the resultant displacement of the delivery truck is approximately 1.6 km, NE from the origin.

3 0

3 years ago

A particle travels clockwise on a circular path of diameter R, monitored by a sensor on the circle at point P; the other endpo

kotykmax [81]

We make a graphic of this problem to define the angle.

The angle we can calculate through triangle relation, that is,

$sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}$

With this function we should only calculate the derivate in function of c

$\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}$

That is the rate of change of $\theta$ .

b) At this point we need only make a substitution of 0 for c in the equation previously found.

$\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}$

Hence we have finally the rate of change when c=0.

6 0

3 years ago

Each of two small non-conducting spheres is charged positively, the combined charge being 40 μC. When the two spheres are 50 cm

Phantasy [73]

Answer:

1.44 x 10⁻⁶ C

Explanation:

$q_{1}$ = charge on one sphere

$q_{2}$ = charge on other sphere

$q$ = Total charge on the two spheres = 40 μC

$q_{1}+ q_{2}$ = $q$

$q_{1}+ q_{2}$ = 40 x 10⁻⁶

$q_{1}$ = (40 x 10⁻⁶) - $q_{2}$ eq-1

$r$ = distance between the two spheres = 50 cm = 0.50 m

$F$ = magnitude of force between the two spheres = 2.0 N

Magnitude of force between the two spheres is given as

$F = \frac{k q_{1} q_{2}}{r^{2}}$

$2.0 = \frac{(9\times 10^{9}) ((40\times 10^{-6}) - q_{2}) q_{2}}{0.50^{2}}$

$q_{2}$ = 1.44 x 10⁻⁶ C

7 0

3 years ago

How many electrons point in 2 second in a wire carrying a current of 20mA ?​

How many electrons point in 2 second in a wire carrying a current of 20mA ?