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Ostrovityanka [42]

3 years ago

13

From the mass spectrum of the compound, the molecule was determined to have a molar mass of 907 g/mol. what is its molecular for

mula?

1 answer:

Molodets [167]3 years ago

3 0

Please elaborate more on your question so I can help you

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The primary service used by stations to exchange mac frames when the frame must traverse the ds to get from a station in one bss

Alenkinab [10]

<span>The primary service used by stations to exchange mac frames when the frame must traverse the ds to get from a station in one bss to a station in another bss is Distribution
In this type of service, the bss is required in order to build the basic building block for the wireless LAN</span>

7 0

3 years ago

Let's take two balls, each of radius 0.1 meters and charge them each to 3 microCoulombs. Then, let's put them on a flat surface

salantis [7]

Explanation:

First, we will calculate the electric potential energy of two charges at a distance R as follows.

R = 2r

= $2 \times 0.1 m$

= 0.2 m

where, R = separation between center's of both Q's. Hence, the potential energy will be calculated as follows.

U = $\frac{k \times Q \times Q}{R}$

= $\frac{8.98 \times 10^{9} \times (3 \times 10^{-6} C)^{2}}{0.1}$

= 0.081 J

As, both the charges are coming towards each other with the same energy so there will occur equal sharing of electric potential energy between these two charges.

Therefore, when these charges touch each other then they used to posses maximum kinetic energy, that is, $\frac{U}{2}$ .

Hence, K.E = $\frac{U}{2}$

= $\frac{0.081}{2}$

= 0.0405 J

Now, we will calculate the speed of balls as follows.

V = $\sqrt{\sqrt{\frac{2 \times K.E}{m}}$

= $\sqrt{\sqrt{\frac{2 \times 0.0405}{4 kg}}$

= 0.142 m/s

Therefore, we can conclude that final speed of one of the balls is 0.142 m/s.

6 0

3 years ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

2 years ago

Where will the spacecraft be when the gravitational forces acting on it are equal?

klio [65]

It would be not be able to move yet it would be in the air

6 0

3 years ago

Dust settling out of the air<br> chemical change or physical change

mars1129 [50]

Physical changes: melting, evaporating, and condensation. This is a physical change.

7 0

3 years ago

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