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sergey [27]
3 years ago
11

At t = 0, one toy car is set rolling on a straight track with initial position 13.0 cm, initial velocity -3.6 cm/s, and constant

acceleration 2.20 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 11.5 cm, initial velocity 5.40 cm/s, and constant zero acceleration. (a) At what time, if any, do the two cars have equal speeds? (Enter NA if the cars never have equal speeds.) s (b) What are their speeds at that time? (Enter NA if the cars never have equal speeds.) cm/s (c) At what time(s), if any, do the cars pass each other? (If there is only one time, enter NA in the second blank. If there are two times, enter the smaller time first. If they never pass, enter NA in both blanks.) s s (d) What are their locations at that time? (If there is only one position, enter NA in the second blank. If there are two positions, enter the smaller position first. If they never pass, enter NA in both blanks.) cm cm (e) Explain the difference between question (a) and question (c) as clearly as possible.
Physics
1 answer:
Klio2033 [76]3 years ago
8 0

Answer:

that's too much to read

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Answer:constant cause it keeps happening Or it might be decreasing but I’m not sure
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What % of an object’s mass is above the water line if the object’s density is 0.82g/ml?
Sladkaya [172]

To develop this problem we will apply the Archimedes model. As well as the definitions of Weight based on mass and acceleration. The first in turn will be considered under the relationship of Density and Volume. From the values given we have to:

\rho_w =1 g/mL \rightarrow \text{Water Density}

\rho_o = 0.82g/mL \rightarrow \text{Object density}

Since it is in equilibrium, the weight of the object will have a reaction from the water, which will cause the sum of forces between the two objects to be zero, therefore

\sum F= 0

F_w-F_o = 0

F_w = F_o

m_w g = m_og

\rho_w V_w g = \rho_o V_o g

The value of gravity is canceled because it is a constant

\frac{V_w}{V_o} = \frac{\rho_o}{\rho_w}

\frac{V_w}{V_o} = \frac{0.82}{1}

\frac{V_w}{V_o} = 0.82

The portion of the object that is submerged corresponds to 82%, while the portion that is visible, above the water level will be 18%

4 0
3 years ago
In preparation for the final exam, our astronomy study group has reconvened to discuss Mercury's unique orbital properties. They
grigory [225]

Answer:

The only incorrect statement is from student B

Explanation:

The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.

Let's examine student claims using these rotation periods

Student A. The time for 4 turns around the sun is

           t = 4 88

           t = 352 / 58.7 Earth days

In this time I make as many rotations on itself each one with a time to = 58.7 Earth days

           #_rotaciones = t / to

           #_rotations = 352 / 58.7

           #_rotations = 6

therefore this statement is TRUE

student B. the planet rotates 6 times around the Sun

          t = 6 88

          t = 528 s

The number of rotations on itself is

           #_rotaciones = t / to

           #_rotations = 528 / 58.7

           #_rotations = 9

False, turn 9 times

Student C. 8 turns around the sun

           t = 8 88

           t = 704 days

the number of turns on itself is

            #_rotaciones = t / to

            #_rotations = 704 / 58.7

            #_rotations = 12

True

The only incorrect statement is from student B

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