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3 years ago

Describe ehat happens at the molecular level during meilting

1 answer:

3 0

The molecules of a solid vibrate faster so that they start spreading out to become a liquid. This energy makes them vibrate faster so the bonds between molecules can't interact all that well anymore creating more distance. The stronger the bonds between the molecules the higher the energy (temperature) has to be to get them away from each other. Hope I didn't confuse you too much!

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lawyer [7]

Answer:

Explanation:

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3 years ago

Read 2 more answers

A long electric cable is suspended above the earth and carries a current of 345 A parallel to the surface of the earth. The eart

jok3333 [9.3K]

Answer:

0.906 N

Explanation:

Formula for magnetic force acting on current carrying cable:

$F = IBLsin(\theta)$

Where I = 345A is the current in the wire, B = $5.6*10^{-5} T$ is the magnetic magnitude generated by Earth. L = 46.9 m is the cable length. $\theta = 88.2^o$ is the angle between vector B and cable direction.

$F = 345*5.6*10^{-5}*46.9*sin(88.2^o)$

$F = 0.906 N$

7 0

3 years ago

A river 1.00 mile wide flows with a constant speed of 1.00 mph. A man can row a boat at 2.00 mph. He crosses the river in a dire

gladu [14]

To solve this problem we will apply the geometric concepts of displacement according to the description given. Taking into account that there is an initial displacement towards the North and then towards the west, therefore the speed would be:

$V_T^2=v_N^2-v_W^2$

$V_T = \sqrt{v_N^2-v_W^2}$

Travel north 2mph and west to 1mph, then,

$V_T = \sqrt{2^2-1^2}$

$V_T = \sqrt{3}$

The route is done exactly the same to the south and east, so make this route twice, from the definition of speed we have to

$v= \frac{\Delta x}{t}$

$t = \frac{\Delta x}{v}$

$t = \frac{2*(1mile)}{\sqrt{3}mph}$

$t = 1.15hour$

Therefore the total travel time for the man is 1.15hour.

3 0

3 years ago

Do anyone k ow the answers?

OLga [1]

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3 years ago

A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci

vovikov84 [41]

According to the law of conservation of momentum:

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.

m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?

$m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'$
$(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'$
$1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')$
$-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')$

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.

$-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')$
$7.5kg.m/s=(235kg)(v_{2}')$
$(7.5kg.m/s)/(235kg)=(v_{2}')$
$0.03m/s=(v_{2}')$

The velocity of the 2nd car after the collision is 0.03m/s.

5 0

3 years ago