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2 years ago

State and briefly describe the applications of expansion

1 answer:

6 0

When solid material expands in response to an increase in temperature (thermal expansion), it can increase in length in a process known as linear expansion. for an example application of expansion and contraction.

examples =
(1) Changing of shape and dimensions of objects such as doors.
(2) Wall collapsing due to bulging.
(3) Cracking of glass tumbler due to heating.
(4) Bursting of metal pipes carrying hot water or steam are some of the disadvantages of thermal expansion of matter.

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WILL MARK AS BRAINLIEST...,.. A constant force vector f =2i cap+3j cap-5k cap acts on a particle and displaces it from (1,2,-3)

Montano1993 [528]

Explanation:

Given that,

Force, $F=2i+3j-5k$

The particle displaces from (1,2,-3) m to (2,5,-1) m.

We need to find the work done by the force. Work done by the force is given by :

W = Fd

It is equal to the dot product of force and displacement.

Displacement from (1,2,-3) m to (2,5,-1) m is (2-1, 5-2, -1-(-3)) or (1, 3, 2) m

Work done,

$W=F{\cdot} d\\\\W=(2i+3j-5k){\cdot} (i+3j+2k)$

We know that, i.i=j.j=k.k=1

So,

$W=1\ J$

So, the work done by the force is 1 J.

6 0

4 years ago

Andrew pokes a marble, and the marble rolls down a ramp. The marble moves with speed. Which forces are acting on the marble in t

stich3 [128]

The forces are Andrew poking the marble and then gravity pulling the marble downward

4 0

3 years ago

Read 2 more answers

Using a radar gun, you emit radar waves at a frequency of 6.2 GHz that bounce off of a moving tennis ball and recombine with the

Keith_Richards [23]

Answer:

23.4 m/s

Explanation:

f = actual frequency of the wave = 6.2 x 10⁹ Hz

$f_{app}$ = frequency observed as the ball approach the radar

$f_{rec}$ = frequency observed as the ball recede away from the radar

V = speed of light

$v$ = speed of ball

B = beat frequency = 969 Hz

frequency observed as the ball approach the radar is given as

$f_{app}=\frac{f(V+v)}{V}$ eq-1

frequency observed as the ball recede the radar is given as

$f_{rec}=\frac{f(V-v)}{V}$ eq-2

Beat frequency is given as

$B = f_{app} - f_{rec}$

Using eq-2 and eq-1

$B = \frac{f(V+v)}{V}- \frac{f(V-v)}{V}$

inserting the values

$969 = \frac{(6.2\times 10^{9})((3\times 10^{8})+v)}{(3\times 10^{8})}- \frac{(6.2\times 10^{9})((3\times 10^{8})-v)}{(3\times 10^{8})}$

$v$ = 23.4 m/s

8 0

3 years ago

Find the force in newtons that does 0.0284 kilojoules of work in moving a book a distance of 4.00 meters

gulaghasi [49]

Work = (force) x (distance

28.4 joules = (force) x (4 meters)

Divide each side by (4 meters) :

Force = (28.4 joules) / (4 meters)

Force = 7.1 Newtons

5 0

4 years ago

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second

Pie

Answer:

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the <u>paths</u> meet, not the point where the particles meet themselves.

So, we can name the time of the first particle $T_F$ , and the time of the second particle $T_S$ .