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3 years ago

If the nucleus of an atom contains 12 protons how many electrons are there in a neutral atom why

1 answer:

7 0

For a neutral atom number of electrons equals number of protons, in other for the net charge of the atom to be zero...
no. of electrons = 12

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guajiro [1.7K]

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3 years ago

Calculate the energy and power required for a student to bike their way to college from home. The student bikes from home for 0.

gregori [183]

Answer:

Check the explanation

Explanation:

To solve the problem, we need to analyze all forces acting on a bicycle individually. In question, student bikes on flat terrain so gravity force doesn't affect the road load. This is the case of uniform acceleration and deceleration so need to calculate average velocity to find Air resistance.

Kindly check the attached images below to see the step by step explanation to the question above.

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3 years ago

I'll give brainliest can someone help me with 2.4.1-2.4.3?with explanation.

KIM [24]

Answer:

220÷4=550 per hour

2.4.3 550×4=2200

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3 years ago

The impulse experienced by a body is equivalent to the body’s change in

nika2105 [10]

<h2>Answer: </h2>

Momentum

<h2>Explanation: </h2>

The momentum of a particle is defined as the product of the particle mass and the particle velocity as follows:

$\overrightarrow{p}=m\overrightarrow{v}$

On the other hand, the impulse of a constant force is defined as:

$\overrightarrow{J}=\varSigma\overrightarrow{F}(t_{2}-t_{1})=\varSigma\overrightarrow{F}\Delta t$

We also know that the net force acting on a particle equals the rate of change of the particle’s momentum, so:

$\varSigma\overrightarrow{F}=m\overrightarrow{a}=m\frac{d}{dt}(\overrightarrow{v})=\frac{d}{dt}(m\overrightarrow{v})=\frac{d\overrightarrow{p}}{dt}$

If the force is constant, then $\frac{d\overrightarrow{p}}{dt}$ equals the total change in momentum over a period of time:

$\varSigma\overrightarrow{F}=\frac{\overrightarrow{p_{2}}-\overrightarrow{p_{1}}}{t_{2}-t_{1}} \\ \\ \varSigma\overrightarrow{F}(t_{2}-t_{1})=\overrightarrow{p_{2}}-\overrightarrow{p_{1}} \\ \\ \boxed{\overrightarrow{J}=\Delta \overrightarrow{p}}$

3 0

3 years ago

A tank whose bottom is a mirror is filled with water to a depth of 20.0 cm. A small fish floats motionless 7.0 cm under the surf

liberstina [14]

Answer:

The apparent depth of (a) the fish is 5.3 cm and (b) the image of the fish is 24.8 cm.

Explanation:

According to the following equation:

$\frac{n_{w} }{s} +\frac{n_{a} }{s'} = \frac{n_{a}- n_{w}}{R_{c} } \\$

where <em>nw</em> and <em>na</em> is the refractive indices of water (1.33) and air (1.00) respectively; <em>s</em> is the depth of the fish below the surface of the water; s' is the apparent depth of the fish from normal incidence and Rc is the radius of curvature of the mirror at the bottom of the tank.

Note that the bottom of the tank is assumed to be a flat mirror, therefore the radius of curvature is very large (R⇒∞).

Therefore, the above equation can be expressed as:

$\frac{n_{w}}{s} +\frac{n_{a}}{s'}=0$

Now we can solve for the apparent depth of the fish.

(a) $s'=-(\frac{n_{a}}{n_{w}})x s$ (Make s' subject of the formula from the above equation)

$s'=(\frac{1.00}{1.33} )x7cm$

∴ $s'=5.3 cm.$

(b) The motionless fish floats 13 cm above the mirror, therefore the image of the fish will be situated at 13 + 20 =33 cm away from the real fish.

Therefore, s = 33 cm

$s'=-(\frac{n_{a}}{n_{w}})x s$

$s'=(\frac{1.00}{1.33} )x 33 cm$

$s'=24.8 cm.$

NB: Here, it is assumed that the water is pure, as impurities may alter the refractive index of water.

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4 years ago