All categories

3 years ago

How many oxygen (O) atoms are in a molecule of SiO2? O A. 1 OB. 3 C. 4 D. 2

Physics

2 answers:

mafiozo [28]3 years ago

3 0

I believe it is D. 2

nikdorinn [45]3 years ago

3 0

The answer is D. 2 because O2 means there is 2 atoms of oxygen.

You might be interested in

A slinky forms it’s third harmonic standing wave when the input frequency is 24 Hz. What is the fundamental frequency of the sli

jarptica [38.1K]

Answer: The fundamental frequency of the slinky = 8Hz

An input frequency of 28 Hz will not create a standing wave

Explanation:

Let Fo = fundamental frequency

At third harmonic,

F = 3Fo

If F = 24Hz

24 = 3Fo

Fo = 24/3 = 8Hz

If an input frequency = 28 Hz at 3rd harmonic

Let find the fundamental frequency

28 = 3Fo

Fo = 28/3

Fo = 9.33333Hz

Since Fo isn't a whole number, it can't create a standing wave

6 0

4 years ago

Buoyancy increases with the increase in the density of a) Submerged body b) Fluid

AlekseyPX

Answer:

Submerged body

Explanation:

If buoyancy is greater than weight then object will float.
If buoyancy is less then weight then object will sink.
If buoyancy=weight then objects remains stable

4 0

3 years ago

A long solenoid with 1.65 103 turns per meter and radius 2.00 cm carries an oscillating current I = 6.00 sin 90πt, where I is in

Leno4ka [110]

Answer:

The electric field is $35\cos(90\pi t)\ mV/m$

Explanation:

Given that,

Radius = 2.00 cm

Number of turns per unit length $n= 1.65\times10^{3}$

Current $I = 6.00\sin 90\pi t$

We need to calculate the induced emf

$\epsilon =\mu_{0}nA\dfrac{dI}{dt}$

Where, n = number of turns per unit length

A = area of cross section

$\dfrac{dI}{dt}$ =rate of current

Formula of electric field is defined as,

$E=\dfrac{\epsilon}{2\pi r}$

Where, r = radius

Put the value of emf in equation (I)

$E=\dfrac{\mu_{0}nA\dfrac{dI}{dt}}{2\pi r}$ ....(II)

We need to calculate the rate of current

$I=6.00\sin 90\pi t$ ....(III)

On differentiating equation (III)

$\dfrac{dI}{dt}=90\pi\times6.00\cos(90\pi t)$

Now, put the value of rate of current in equation (II)

$E=\dfrac{4\pi\times10^{-7}\times1.65\times10^{3}\times\pi\times(2.00\times10^{-2})^2\times90\pi\times6.00\cos(90\pi t)}{2\pi\times 2.00\times10^{-2}}$

$E=35\cos(90\pi t)\ mV/m$

Hence, The electric field is $35\cos(90\pi t)\ mV/m$

7 0

3 years ago

Read 2 more answers

Li and Raj are playing football. They take turns trying to run past each other with the football. Li weighs 125 pound and Raj we

Elenna [48]

Answer:

Li has less mass and therefore less inertia, so he can change his motion more easily than Raj.

Explanation:

Inertia describes the resistance of an object to any change in its state of motion, and it depends on the mass of the object only. In particular:

- if an object has a large inertia (large mass), then it is more difficult to change its state of motion (i.e. to put it in motion, or to slow it down, or to change its direction of motion)

- if an object has small inertia (small mass), then it is more easy to change its state of motion

In this problem, Li has less mass than Raj, so he has less inertia, therefore he can change his motion more easily than Raj.

3 0

4 years ago

Read 2 more answers

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$

5 0

3 years ago