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2 years ago

How many oxygen atoms are in Ca (NO3)2

Physics

1 answer:

Lapatulllka [165]2 years ago

5 0

There are 6 atoms of oxygen in Ca (NO3)2.

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9. A current of 9 A flows through an electric device with a resistance of 43 Ω. What must be the applied voltage in this particu

kotykmax [81]

Voltage =Current * Resistance

= 9 * 43

Voltage = 387 V

6 0

3 years ago

A light ray incident on a block of glass makes an incident angle of 50.0° with the normal to the surface. The refracted ray in t

Andru [333]

Answer:

$n_{glass} = 1.65$

Explanation:

As we know that the angle of incidence is given as

$i = 50.0^o$

also we have angle of refraction as

$r = 27.7^o$

now by Snell's law we know that

$n_{air} sin i = n_{glass} sin r$

$1 sin50 = n_{glass} sin 27.7$

now we have

$n_{glass} = \frac{sin 50}{sin 27.7}$

$n_{glass} = 1.65$

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3 years ago

Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l

OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

$Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\$

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

$F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\$

8 0

2 years ago

What is the equivalent resistance of the circuit?

Ivanshal [37]

Answer:

Since the wire is not splitting at any point in the circuit,

the resistors are in series

Hence, Equivalent resistance = 10 + 20 + 30

Equivalent Resistance = 60 Ω

8 0

3 years ago

The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?

vitfil [10]

The expression of the electric flux is

$\Phi = \frac{Q}{\epsilon_0}$

Here,

Q = Total charge enclosed in the closed surface

$\epsilon_0$ = Permittivity due to free space

Rearranging to find the charge,

$Q = \epsilon_0 \Phi$

Replacing with our values we have finally

$Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)$

$Q = 1.6284*10^{-8} C$ $(\frac{10^9nC}{1C})$

$Q = 0.1684nC$

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0

2 years ago