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What is the kinetic energy of a proton that is traveling at a speed of 3725 m/s ?

Short Answer

Expert verified

The kinetic energy of a proton is 1.16×10-20 J.

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The travelling speed of a proton is, v=3725 m/s.
02

Significance of kinetic energy

The kinetic energy of a body mainly depends on two physical quantities: speed and mass of the body. If the speed of a heavy (large mass) body is more, then the body consists of large kinetic energy.

03

Determination the kinetic energy of a proton 

The relation of kinetic energy of a proton is expressed as,

(KE)p=12mpv2

Here,(KE)pis the kinetic energy of a proton, and mpis the mass of a proton whose value is 1.67×10-27 kg.

Substitute all the known values in the above equation.

(KE)p=121.67×10-27kg3725m/s2≈1.16×10-20kg.m2/s2≈1.16×10-20kg.m2/s2×1J1kg.m2/s2≈1.16×10-20J

Thus, the kinetic energy of the proton is 1.16×10-20 J.

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Most popular questions from this chapter

Question: In Chapter 6 we saw that the electric potential energy of a system of two particles is given by the equation Uel=14πε0q1q2r.

(a) What is the electric potential energy of two protons separated by a distance of 9nm9×10-9m? (b) What is the electric potential energy of a proton and an electron separated by the same distance?

An electron is initially at rest. It is moved from a location 4×10-10mfrom a proton to a location 6×10-10m from the proton. What is the change in electric potential energy of the system of proton and electron?

A dipole is centered at the origin, with its axis along the y axis, so that at locations on the y axis, the electric field due to the dipole is given by

E→=⟨0, 14πε02qsy3, 0⟩ Vm

The charges making up the dipole are+3 n°ä and -3 n°ä, and the dipole separation is2″¾³¾ (Figure 16.82). What is the potential difference along a path starting at location⟨0, 0.03, 0⟩″¾ and ending at location⟨0, 0.04, 0⟩″¾ ?

An isolated large plate capacitor (not connected to anything) originally has a potential difference of 1000 V with an air gap of 2 mm. Then a plastic slab 1 mm thick, with dielectric constant 5, is inserted in the middle of the air gap as shown in figure 16.96. Calculate the following potential differences and explain your work.

V1-V2=?V2-V3=?V3-V4=?V1-V4=?

In the region of space depicted in Figure 16.87 there are several stationary charged objects that are not shown in the diagram. Along a path A=B=C=D you measure the following potential differences:VB−VA=12 V ;VC−VB=−5 V ; VD−VC=−15 V. What is the potential differenceVA−VD ?

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