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3 years ago

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A particle (m = 4.3 × 10^-28 kg) starting from rest, experiences an acceleration of 2.4 × 10^7 m/s^2 for 5.0 s. What is its de B

roglie wavelength λ at the end of this period?

1 answer:

Novay_Z [31]3 years ago

4 0

Answer:

Wavelength, $\lambda=1.28\times 10^{-14}\ m$

Explanation:

Given that,

Mass of the particle, $m=4.3\times 10^{-28}\ kg$

Acceleration of the particle, $a=2.4\times 10^7\ m/s^2$

Time, t = 5 s

It starts from rest, u = 0

The De Broglie wavelength is given by :

$\lambda=\dfrac{h}{mv}$

v = a × t

$\lambda=\dfrac{h}{mat}$

$\lambda=\dfrac{6.67\times 10^{-34}}{4.3\times 10^{-28}\times 2.4\times 10^7\times 5}$

$\lambda=1.28\times 10^{-14}\ m$

Hence, this is the required solution.

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When we get out of the bed on a very cold morning, we feel that the air of the room is cold. But when we come back after staying

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Answer:

This is because the air outside is always cooler than the air inside, so after staying outside your body adapts to the cold air, when you come back inside, the cold air is still in you which makes the room seem warmer.

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Find the moment of inertia Ihoop of a hoop of radius r and mass m with respect to an axis perpendicular to the hoop and passing

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Answer: MR²

is the the moment of inertia of a hoop of radius R and mass M with respect to an axis perpendicular to the hoop and passing through its center

Explanation:

Since in the hoop , all mass elements are situated at the same distance from the centre , the following expression for the moment of inertia can be written as follows.

I = ∫ r² dm

= R²∫ dm

MR²

where M is total mass and R is radius of the hoop .

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3 years ago

An object moving in circular motion has a mass of 15 kg and a centripetal acceleration of 10 m/s2. What is the centripetal force

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Answer:

1) A

2) C

3) B

4) A

5) Incomplete information(picture missing)

6) Incomplete information(picture missing)

7) Incomplete information(picture missing)

8) A

9) C

10) C

Explanation:

1) m = 15kg, a = $10ms^{-2}$ , F = ma = 15*10 = 150N

2) m = 3kg, v = $4ms^{-1}$ , r = 4m, F = $\frac{mv^{2} }{r}$

$\frac{3*4^{2} }{4}$ = 12N

3) a = $10ms^{-2}$ , r = 10m, v=?

F = $\frac{mv^{2} }{r}$ and F = ma

equating the two equations and cancelling a, we have:

$\frac{v^{2} }{r}$ = a

making v the subject of formula, we have:

v = $\sqrt{ar}$

= $\sqrt{100}$

= 10 $ms^{-1}$

4) r = 10m, v = $5ms^{-1}$ , a = ?

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

F = ma

equating the above equations and making a subject of formula, we get:

a = $\frac{v^{2} }{r}$

a = 25/10 = 2.5 $ms^{-2}$

5) I can't find the picture associated with this question

6) I can't find the picture associated with this question

7) I can't find the picture associated with this question

8) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is tripled

F = $(3v)^{2}$

F = 9 $v^{2}$

We can see that the force will be 9X greater than it was.

9) F = $\frac{mv^{2} }{r}$

assuming m and r is unity, that is the values are 1 respectively, the formula simplifies to:

F = $v^{2}$

Now, if v is doubled

F = $(2v)^{2}$

F = 4 $v^{2}$

We can see that the force will be 4X greater than it was.

10) F = $\frac{mv^{2} }{r}$

assuming m and v is unity, that is the values are 1 respectively

F = 1/r

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F = 1/2 * 1/r

We can see that the force is 1/2 as big as it was

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Oxidation reactions in biological systems generally involve

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Answer: addition of electrons and hydrogen ions. loss of electrons and addition of hydrogen ions. loss of electrons and either addition or loss of hydrogen ions.

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3 years ago

A person throws a ball straight up. He releases the ball at a height of 1.75 m above the ground and with a velocity of 12.0 m/s.

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Answer:

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b) 9.089 m

Explanation:

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u = Initial velocity

v = Final velocity

s = Displacement

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$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-12}{-9.81}\\\Rightarrow t=1.22\ s$

Time taken by the ball to reach the highest point is 1.22 seconds

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The maximum height the ball will reach above the ground is 1.75+7.339 = 9.089 m

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