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Karo-lina-s [1.5K]

2 years ago

12

If you are at latitude 43 degrees north of Earth's equator, what is the angular distance (in degrees) from your zenith to the no

rth celestial pole?

1 answer:

kirill115 [55]2 years ago

8 0

Answer:

Your zenith is 43 N of 90 deg (equator)

Thus, your zenith is 90 - 43 = 47 deg

(At the N pole your zenith would be 0 deg from the N pole)

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Set the radius to 2.0 m and the velocity to 1.0 m/s. Keeping the radius the same, record the magnitude of centripetal accelerati

jek_recluse [69]

Answer:

$a=4\ m/s^2$

Explanation:

Given that,

Radius, r = 2 m

Velocity, v = 1 m/s

We need to find the magnitude of the centripetal acceleration. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{1}\\\\=4\ m/s^2$

So, the magnitude of centripetal acceleration is $4\ m/s^2$ .

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2 years ago

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3 years ago

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A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel

zloy xaker [14]

Answer:

$4.25 m/s^{2}$

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

$\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s$

Change in velocity considering the y component will be

Final velocity-Initial velocity

$\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s$

Resultant change in velocity $=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s$

Acceleration= change in velocity per unit time hence

$a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}$

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3 years ago

If Calcium has 2 valence electrons and Chlorine has 7 valence electrons, how many Cl-1 ions would need to combine with 1 Ca+2 io

Tanya [424]

Answer:

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Explanation:

6 0

2 years ago

A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T th

N76 [4]

The mass of the ion is 5.96 X 10⁻²⁵ kg

<u>Explanation:</u>

The electrical energy given to the ion Vq will be changed into kinetic energy $\frac{1}{2}mv^2$

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force $\frac{mv^2}{r}$ .

So,

$Vq = \frac{1}{2}mv^2$

and

$Bqv = \frac{mv^2}{r}$

Right from these eliminating v, we can derive

$m = \frac{B^2r^2q}{2V}$

On substituting the value, we get:

$m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\$

m = 5.96 X 10⁻²⁵ kg.

3 0

3 years ago