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Effectus [21]
3 years ago
8

What is the kinetic energy of a 24-kg mass that is moving with a velocity of 2 m/s?

Physics
2 answers:
Anton [14]3 years ago
3 0

Answer:

1/2mv^2

1/2*24*2^2

answer=48

Alina [70]3 years ago
3 0

Answer:

48 J

Explanation:

Kinetic energy can be found using the following formula:

KE=\frac{1}{2}mv^2

where <em>m </em>is the mass in kilograms and <em>v </em>is the velocity in m/s.

We know the mass is 24 kilograms and the velocity is 2 m/s Therefore, substitute 24 in for <em>m </em>and 2 in for <em>v. </em>

<em />KE=\frac{1}{2}*24*2^2<em />

Evaluate the exponent first. 2^2 is the same as multiply 2, 2 times.

2^2=2*2=4

KE=\frac{1}{2}*24*4

Multiply 24 and 4

KE=\frac{1}{2} *96

Multiply 96 and 1/2, or divide 96 by 2.

KE=48

Add appropriate units. Kinetic energy uses Joules, or J

KE=48 Joules

The kinetic energy is 48 Joules

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Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

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3 years ago
A farmer hitches her tractor to a sled loaded with firewood and pulls it a distance
Delvig [45]

(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

<h3>Work done by the applied force</h3>

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

<h3>Work done by frictional force</h3>

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

<h3>Net work done by all the forces on the tractor</h3>

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

Learn more about work done here: brainly.com/question/25573309

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