answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
liq [111]
2 years ago
10

A metal bar is elongated by 2mm when put under a certain amount of tension. How much will it be elongated if the tension is doub

led?
Physics
1 answer:
Inessa [10]2 years ago
5 0

Answer: 4mm

Explanation: From Hooke's law.

Tensile force is directly proportional to the elongation

That's

F= k e , where k is constant

F/e = k

F1/e1 =F2/e2

Where e1 = 2mm

F2 =2F1 (doubled)

F1/2 = 2F1/e2

1/2 =2/e2

e2 = 4mm

Therefore when tension is doubled, the metal bar will be elongated by 4mm

You might be interested in
Physics in motion unit 6a the nature of waves
mylen [45]
What’s the question?
7 0
2 years ago
Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f
stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2

When s = Db, t = 2t

s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2

Dividing the two equations

\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db

Hence, Da=(1/4)Db

3 0
2 years ago
A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10
alisha [4.7K]
(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
I= \frac{1}{2} c \epsilon_0 E^2
where
I is the intensity
c is the speed of light
\epsilon_0 is the electric permittivity
E is the amplitude of the electric field

By substituting the numbers of the problem and re-arranging the equation, we can find E:
E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C

Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
F=qE=(2.1 \cdot 10^{-8} C)(2.2 \cdot 10^6 N/C)=0.046 N

(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T

The magnetic force is given by
F=qvB \sin \theta
where v is the particle's speed, B the magnetic field intensity and \theta the angle between B and v.
In this case the charge is stationary, so v=0, and so the magnetic force is zero: F=0.

(c) The electric force has not changed compared to point (a), because it does not depend on the speed of the particle, so we have again F=0.046 N.

(d) This time, the particle is moving with speed v=3.7 \cdot 10^4 m/s, in a direction perpendicular to the magnetic field (so, the angle \theta is 90^{\circ}), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=
=5.7 \cdot 10^{-6} N
5 0
2 years ago
calculate the workdone to stretch an elastic string by 40cm if a force of 10N produces an extension of 4cm in it
Lera25 [3.4K]
The force of F=10 N produces an extension of
x=4 cm=0.04 m
on the string, so the spring constant is equal to
k= \frac{F}{x}= \frac{10 N}{0.04 m}=250 N/m

Then the string is stretched by \Delta x=40 cm=0.40 m. The work done to stretch the string by this distance is equal to the variation of elastic potential energy of the string with respect to its equilibrium position:
W= \Delta U= \frac{1}{2}k(\Delta x)^2  = \frac{1}{2}(250 N/m)(0.40 m)^2=20 J
5 0
2 years ago
A physics student shoves a 0.50-kg block from the bottom of a frictionless 30.0° inclined plane. The student performs 4.0 j of w
Musya8 [376]

Answer:1.63 m

Explanation:

Given

mass of block m=0.5 kg

inclination \theta =30^{\circ}

Amount of work done W=4 J

block slides a distance s along the Plane

Work done =change in Potential Energy

Increase in height of block is s\sin \theta

Change in Potential Energy =mg(\sin \theta -0)

\Delta P.E.=0.5\times 9.8\times s\sin 30

4=0.5\times 9.8\times s\sin 30

s=\frac{4}{2.45}      

s=1.63 m

5 0
2 years ago
Other questions:
  • A 1,300 kg wrecking ball hits the building at 1.07 m/s2.
    11·2 answers
  • An 8.0-kg history textbook is placed on a 1.25-m high desk. How large is the gravitational potential energy of the textbook-Eart
    11·2 answers
  • The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene
    5·2 answers
  • In a physics laboratory experiment, a coil with 200 turns enclosing an area of 12cm2 is rotated in 0.040 s from a position where
    7·1 answer
  • This is a cell, which is the basic unit of all life. All organs in human bodies are made of cells and require oxygen to survive.
    5·1 answer
  • 4 A wheel starts from rest and has an angular acceleration of 4.0 rad/s2. When it has made 10 rev determine its angular velocity
    13·1 answer
  • Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio
    14·2 answers
  • Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp
    7·1 answer
  • Sophia was studying with Maalik about the layers of the Earth. They were given several different statements about the layers. Wh
    12·2 answers
  • A boy jumps into an indoor swimming pool. He notices that the water appears to get colder as he
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!