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2 years ago

Iron's ability to rust is not a physical property because

Physics

2 answers:

gtnhenbr [62]2 years ago

6 0

The answer should be c

butalik [34]2 years ago

4 0

C. a new substance with new properties is formed.

A physical property is a property that can be observed and what not, but does not change the original object/form a new substance, hence why the answer is C

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A 220 kg crate hangs from the end of a rope of length L = 14.0 m. You push horizontally on the crate with a varying force F to m

kifflom [539]

Answer:

(a) magnitude of F = 797 N

(b)the total work done W = 0

(c)work done by the gravitational force = -1.55 kJ

(d)the work done by the pull = 0

(e) work your force F does on the crate = 1.55 kJ

Explanation:

Given:

Mass of the crate, m = 220 kg

Length of the rope, L = 14.0m

Distance, d = 4.00m

(a) What is the magnitude of F when the crate is in this final position

Let us first determine vertical angle as follows

=> $Sin \theta = \frac{d }{L}$

=> $\theta = Sin^{-1} \frac{d}{L}$ =

Now substituting thje values

=> $\theta = Sin^{-1} \frac{4}{12}$ =

=> $\theta = Sin^{-1} \frac{1}{3}$

=> $\theta = Sin^{-1}(0.333)$

=> $\theta = 19.5^{\circ}$

Now the tension in the string resolve into components

The vertical component supports the weight

=> $Tcos\theta = mg$

=> $T = \frac{mg}{cos\theta}$

=> $T = \frac{230 \times 9.8 }{cos(19.5)}$

=> $T = \frac{2254 }{cos(19.5)}$

=> $T = \frac{2254 }{0.9426}$

=>T =2391N

Therefore the horizontal force

$F = TSin(19.5)$

F = 797 N

b) The total work done on it

As there is no change in Kinetic energy

The total work done W = 0

c) The work done by the gravitational force on the crate

The work done by gravity

Wg = Fs.d = - mgh

Wg = - mgL ( 1 - Cosθ )

Substituting the values

= $-230 \times 9.8\times 12 ( 1 - cos(19.5) )$

= $-230 \times 9.8\times 12 ( 1 - 0.9426) )$

= $-230 \times 9.8\times 12 (0.0574)$

= $-230 \times 9.8\times 0.6888$

= $-230 \times 6.750$

= -1552.55 J

The work done by gravity = -1.55 kJ

d) the work done by the pull on the crate from the rope

Since the pull is perpendicular to the direction of motion,

The work done = 0

e)Find the work your force F does on the crate.

Work done by the Force on the crate

WF = - Wg

WF = -(-1.55)

WF = 1.55 kJ

(f) Why is the work of your force not equal to the product of the horizontal displacement and the answer to (a)

Here the work done by force is not equal to F*d

and it is equal to product of the cos angle and F*d

So, it is not equal to the product of the horizontal displacement and the answer to (a)

7 0

3 years ago

Fill in the blanks to complete the sentence.

mina [271]

The answer is:

Fill in the blanks to complete the sentence.

Light acts like a PARTICAL when it bounces off surfaces,

and acts like a WAVE when it bends around objects.

I hope this helps you :>

4 0

3 years ago

Read 2 more answers

The smallest level of organization in living things is

Advocard [28]

The smallest level of organization in living things is the atom.

Next in line would be the cell, since a cell is made up of atoms working together. Next, cells working together would make up a tissue, and further, tissues working together would make up an organ.

6 0

3 years ago

B. If we drop a ball from the roof, its falls downward

frosja888 [35]

Answer:

yes ,what's the question

and if the question is why

then it's for gravity

4 0

2 years ago

Read 2 more answers

A car of mass m goes around a banked curve of radius r with speed v. If the road is frictionless due to ice, the car can still n

Sholpan [36]

Answer:

horizontal component of normal force is equal to the centripetal force on the car

Explanation:

As the car is moving with uniform speed in circle then the force required to move in the circle is towards the center of the circle

This force is due to friction force when car is moving in circle with uniform speed

Now it is given that car is moving on the ice surface such that the friction force is zero now

so here we can say that centripetal force is due to component of the normal force which is due to banked road

Now we have

$N sin\theta = \frac{mv^2}{R}$

$N cos\theta = mg$

so we have

$v = \sqrt{Rg tan\theta}$

so this is horizontal component of normal force is equal to the centripetal force on the car

5 0

3 years ago