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2 years ago

Take the mechanical equivalent of heat as 4 J/cal. A 10-g bullet moving at 2000 m/s plunges

1 answer:

4 0

Assuming that all the bullet’s energy heats the paraffin, its final temperature is 27.1 degree C. The correct option is D.

<h3>What is temperature?</h3>

Temperature is the degree of hotness and coldness of the material.

The energy of the bullet E = 1/2 mv²

E = 1/2 x 10 x 10⁻³ x (2000)²

E = 2 x 10⁴ J

This heat is used in heating the paraffin

E = m x c ΔT = m x c (Tfinal -Tinitial)

2 x 10⁴ J = 1 x 2.8 x 10³ x (Tfinal -20)

Tfinal = 27.1°C

Thus, the final temperature is 27.1 degree C. The correct option is D.

Learn more about temperature.

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A research vessel is mapping the bottom of the ocean using sonar. It emits a short sound pulse called "ping" downward. The frequ

Svetradugi [14.3K]

Answer:

d = 4180.3m

wavelengt of sound is 0.251m

Explanation:

Given that

frequency of the sound is 5920 Hz

v=1485m/s

t=5.63s

let d represent distance from the vessel to the ocean bottom.

an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.

$velocity=\frac{distance}{time}\\\\ v=\frac{2d}{t} \\\\vt=2d\\\\d=\frac{vt}{2}$

$d=\frac{1485*5.63}{2}\\d= 4180.3m$

wavelengt of sound is $\lambda$ = v/f

= (1485)/(5920)

= 0.251 m

7 0

3 years ago

A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig

vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

$CPR=\frac{KW}{DAT}$

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6 to give CPR in mm/yr

mpy

$CPR=\frac{KW}{DAT}$

$=\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}$

=37.4mpy

mm/yr

$CPR=\frac{KW}{DAT}$

$=\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}$

=0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0

3 years ago

What is the force felt by the electrons and the nuclei in the rod when the external field described in the problem introduction

fomenos

Complete Question

The complete question is shown on the uploaded image

Answer:

The nuclei experience a force that will move it to the right of the conductor rod while the electrons experience a force that will move it to the left side of the conductor rod.

Explanation:

The force that act on the charges(both the positive and the negative charge ) is Mathematically expressed as

F = qE => E = F/q

where F is the force, q is the charge and E is the electric field

we can see that the force is directly proportional to the electric field,so an increase in the electric field would increase the force.

we can also see that the electric field is given as force per unit charge and generally the direction of this field is taken to be the direction of the force it would exert on a positive test charge

Now from the question we are being told that the external electric field is the direction of the positive x axis

Hence this field would drive the positive charge i.e the nuclei to the right.

In order to further explain let consider this

Generally the electric field is always radially outward originating from a positive point charge and radially in toward a negative point charge.

what this means for this question, is that the positive point charge is on the left side of the electric field while the negative point charge is at the right side of the field.

According to Coulomb's law which states that unlike term attract while like terms repel, it means that the electron would move to the left of the conductor rod while the nuclei would move to the right side of the conductor rod.