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Lilit [14]
2 years ago
5

When plastic deformation of a material occurs, the material _____. regains its original shape when the stress is removed is perm

anently deformed becomes magnetic at low temperatures becomes highly resistant to stress

Physics
2 answers:
LuckyWell [14K]2 years ago
7 0
Answer: Is permanently deformed.

Explanation (see the diagram shown below).
For most structural materials:
(a) The deformation in the elastic zone is recoverable, and the material returns
      to its original shape when the load (or stress) is removed.
(b) A permanent set (non-recoverable deformation) occurs when the material
     is sressed beyond the elastic zone, that is, in the plastic zone.

ki77a [65]2 years ago
3 0
The answer is permanetly deformed

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Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


5 0
2 years ago
If i apply 280 n of force to a 40kg object, what will it's acceleration be?
nika2105 [10]

Answer:

f=ma

f=280N

m=40kg

a=?

280=40a

a=280/40

a=70N/kg

6 0
3 years ago
for much of the year this desert is hot dry and windy.What type of plant would you most expect to survive in this climate
abruzzese [7]
I think the answer is cactus
7 0
3 years ago
An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas
Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

F_f=\mu mg

where

\mu is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

\mu=0.500 (coefficient of friction)

m = 1260 kg (mass of the car)

g=9.8 m/s^2

Therefore, the force of friction is

F_f=(0.500)(1260)(9.8)=6174 N

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

F=m\frac{v^2}{r}

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

v=54.1 km/h =15.0 m/s is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

F=(1260)\frac{15.0^2}{41.6}=6814 N

Therefore, the force of friction is not enough to keep the car in the curve, since F_f

4 0
3 years ago
Suppose you first walk 12.0 m in a direction 200 west of north and then 20.0 m in a direction 40.00 south of west. How far are y
Tpy6a [65]

Complete Question

The  complete question is shown on the first uploaded image

Answer:

the compass direction of the resultant displacement is  \theta  =4.7^o south of west

Explanation:

Generally using cosine we can obtain the resultant R as follows

     R^2  =  A^2  + B^2 -2ABcos(70)

=>   R  =  \sqrt{12^2  + 20^2  - 2(12 ) *  (20) cos  70}

=>    R  =  19.48 \  m

We can obtain the direction of the resultant by first  using sine rule to obtain angle C as follows

       \frac{A}{sin  C}  =  \frac{R}{sin70 }

=>    C=  sin ^{-1} [\frac{A *  (sin 70)}{R} ]

=>    C =  sin ^{-1} [\frac{20 *  (sin 70)}{19.48} ]

=>  C =  74.7 ^o

Then the direction is obtained as

       \theta  =  C  -  70

=>    \theta  = 74.7   -  70

=>     \theta  =4.7^o

Hence the compass direction of the resultant displacement is  \theta  =4.7^o south of west

7 0
3 years ago
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