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4 years ago

an object moving with a speed of 5 m/s has a kinetic energy of 100 J. what is the mass of the object?

Physics

1 answer:

elixir [45]4 years ago

5 0

Answer:40kg

Explanation:

Velocity(v)=5m/s

Kinetic energy(ke)=100j

mass=?

Mass=(2xke) ➗ v^2

Mass=(2x500) ➗ 5^2

Mass=1000 ➗ 5x5

Mass=1000 ➗ 25

Mass=40kg

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A person shuffles across the rug and then place his finger near his friend's nose, causing a small spark that transfers about 10

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Answer:

$n = \frac{10^{-9}C}{1.6 x10^{-19}C}= 6.25 x10^{9} electrons \approx 6.3 x10^{9} electrons$

Explanation:

For this case we know that the charge associated to the transferred electrons of the small spark is $10^{-19}C$ and we want to find the number of electrons transferred.

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$q_e = |-1.6 x10^{-19}C |= 1.6 x10^{-19}C$

And we know that the total charge is given by this expression:

$Q = n e$

Where Q represent the charge, n the number of electrons and e the charge of the electron.

If we solve for n we got:

$n = \frac{Q}{e}$

And replacing we got:

$n = \frac{10^{-9}C}{1.6 x10^{-19}C}= 6.25 x10^{9} electrons \approx 6.3 x10^{9} electrons$

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3 years ago

An electron is located on a pinpoint having a diameter of 3.52 µm. What is the minimum uncertainty in the speed of the electron?

Komok [63]

Explanation:

It is given that,

An electron is located on a pinpoint having a diameter of 3.52 µm, $\Delta x=3.52\times 10^{-6}\ m$

We need to find the minimum uncertainty in the speed of the electron. It can be calculated using Heisenberg uncertainty principal as :

$\Delta p.\Delta x \geq \dfrac{h}{4\pi}$

$\Delta p \geq \dfrac{h}{4\pi \Delta x}$

Since, p = m v

So, $mv \geq \dfrac{h}{4\pi \Delta x}$

$\Delta v \geq \dfrac{h}{4\pi \Delta x m}$

$\Delta v \geq \dfrac{6.67\times 10^{-34}}{4\pi 3.52\times 10^{-6}\times 9.1\times 10^{-31}}$

$\Delta v\geq 16.57\ m/s$

So, the minimum uncertainty in the speed of the electron is greater than 16.57 m/s. Hence, this is the required solution.

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3 years ago

Two out of phase loudspeakers are some distance apart. A person stands 4.50 m from one speaker and 3.80 m from the other. What i

Contact [7]

Answer:

The third lowest frequency is 1489.27 Hz.

Explanation:

Given that,

Distance of person from one speaker= 4.50 m

Distance of person from the other = 3.80 m

The longest wavelength is the shortest frequency.

We need to calculate the difference between the speakers

The difference between the speakers d= 4.50-3.80 =0.7 m

We know that,

The longest wavelength is such that one wavelength

$\lambda = d$

So, The third longest wavelength is

$3\lambda=d$

Put the value of d

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We need to calculate the lowest frequency

Using formula of frequency

$f=\dfrac{v}{\lambda}$

$f=\dfrac{347}{0.233}$

$f=1489.27\ Hz$

Hence, The third lowest frequency is 1489.27 Hz.

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3 years ago