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user100 [1]
3 years ago
8

Cool the matter back down to a solid state. How does its shape compare to its original shape?Why? *

Physics
1 answer:
tino4ka555 [31]3 years ago
7 0

Answer:

i am just answering what this guy said this answer isnt mine ok -v-/ Both soaps and emulsifiers contain only one nonpolar tail group Both soaps and emulsifiers consist of fatty acid chains and a polar head group Both soaps and emulsifiers work in the same manner

Soaps contain two polar tail groups, but emulsifiers have only one polar tail group ,Soaps contain only one nonpolar tail group, but emulsifiers have two nonpolar tail group

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Alina [70]

Answer:

take daily showers . eat vegies .sanitize your hands

7 0
2 years ago
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A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex
Lynna [10]

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

v_i = v_o

now at the end of 80 s let the speed is

v_f = v_1

after another 120 s let the speed will be

v_f' = v_2

now we know that

d = \frac{v_i + v_f}{2} (t)

d = \frac{v_o + v_1}{2}(80)

1000 = 40(v_o + v_1)

also we know that

v_1 - v_o = a(80)

also we have

1000 = \frac{v_1 + v_2}{2}(120)

1000 = 60(v_1 + v_2)

now we can say

(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}

also we know

v_2 - v_o = a(120 + 80)

-8.33 = 200 a

a = -0.042 m/s^2

Part b)

now we have

v_1 + v_o = 25

v_1 - v_o = (-0.042)(80)

v_1 = 10.83 m/s

so the starting velocity of the trip is

v_o = 25 - 10.83 = 14.17 m/s

now speed after t = 200 s is given as

v_2 = v_o + at

v_2 = 14.17 - (0.042)(200)

v_2 = 5.77 m/s

5 0
3 years ago
How do the three types of boundary's work together to keep the Earth at equilibrium?
Maurinko [17]
There are three types: divergent, convergent, and transform boundaries. I hope this helps.
8 0
2 years ago
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Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire
Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

F = ma \rightarrow a = \frac{F}{m}

Replacing,

a =\frac{42N}{83000kg}

a =5.06*10^{-4}m/s^2

The total speed change

\Delta v = v_f -v_0 \rightarrow v_f =\text{Final velocity and } v_0 = \text{Initial velocity } we have that the value is 0.71m/s

If we know that acceleration is the change of speed in a fraction of time,

a= \frac{\Delta v}{t} \rightarrow t = \frac{\Delta v}{a}

We have that,

t= \frac{0.71m/s}{5.06*10^{-4}m/s^2 }

t = 1403.16s

Therefore the Rocket should be fired around to 1403.16s

7 0
2 years ago
How is constant acceleration indicated on a motion map?
sasho [114]
<span>by vectors that are all the same length

</span>
6 0
2 years ago
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