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Murljashka [212]
3 years ago
8

According to the impulse--momentum equation Ft = change in mv, a bungee jumper in fall has momentum which is reduced by the forc

e F exerted by the bugle cord. If m is the mass of the jumper, then v in the equation is the speed of thea. Jumper
b. Cord
c. Both
d. None
Physics
1 answer:
Dima020 [189]3 years ago
6 0

Answer:

The answer is a. Jumper

Explanation:

The change in the momentum is represented by the equation

Ft=m*v

Acceleration of any body is calculated by the product of the mass of the body and with the velocity the same mass is moving.

Since it is mentioned in the statement that m is the mass of the jumper then velocity will also be of the jumper to calculate the change in the momentum.

If the mass of bugle cord is taken, then its velocity will be taken for the equation.

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A long iron bar lies along the x-axis and has current of I = 16.4 A running through it in the +x-direction. The bar is in the pr
Gre4nikov [31]

Answer:

B = 8.0487mT

Explanation:

To solve the exercise it is necessary to take into account the considerations of the Magnetic Force described by Faraday,

The magnetic force is given by the formula

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According to our data we have that

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F = 0.132N/m

As we know our equation must be modificated to Force per length unit, that is

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8 0
4 years ago
1. An electron in an atom absorbs a photon with an energy of 2.38 eV and jumps from the n = 2 to n = 4 energy level in the atom.
oksian1 [2.3K]

1. 1.0\cdot 10^{-6}m

First of all, let's convert the energy of the absorbed photon into Joules:

E=2.38 eV \cdot (1.6\cdot 10^{-19}J/eV)=1.98\cdot 10^{-19} J

The energy of the photon can be rewritten as:

E=\frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda is the wavelength of the photon

Re-arranging the formula, we can solve to find the wavelength of the absorbed photon:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1.98\cdot 10^{-19} J}=1.0\cdot 10^{-6}m

2. 1.24 eV

In this case, when the electron jumps from the n=4 level to the n=3 level, emits a photon with wavelength

\lambda=1.66\cdot 10^{-6}m

So the energy of the emitted photon is given by the formula used previously:

E=\frac{hc}{\lambda}

and using

\lambda=1.66\cdot 10^{-6}m

we find

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.0\cdot 10^{-6}m}=1.99\cdot 10^{-19}J

converting into electronvolts,

E=\frac{1.99\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=1.24eV

EDIT: an issue in Brainly does not allow me to add the last 2 parts of the solution - I have added them as an attachment to this post, check the figure in attachment.

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