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givi [52]
2 years ago
14

When driving a Toyota avensis car along a straight road for 16.5km at

Physics
1 answer:
Vitek1552 [10]2 years ago
7 0

The average velocity of the whole journey will be total distance covered divided by the total time. It will be approximately equal to 8 m/s. The right answer is option B

<h2>VELOCITY</h2>

Velocity is the distance travelled in a specific direction. While the average velocity of the whole journey will be total distance covered divided by the total time

When driving a Toyota avensis car along a straight road for 16.5km at

50km/h,

The velocity = 50 km/h

Distance = 16.5 km

Use the speed formula to calculate time.

Speed = distance / time

Time = distance / speed

Time = 16.5 / 50

Time = 0.33 s

If over the next 20min, you walked another 2.5km further along the road for a petrol station, Then,

average velocity = Total distance covered divided by total time taken.

Where

The time t = 20/60 = 0.333 h

Total time = 0.33 + 0.3333

Total time = 0.6633333

Total distance = 16.5 + 2.5

Total distance = 19 km

Average velocity = 19 / 0.66333

Average Velocity = 28.64 km/h

Now convert Km/h to m/s

(28.6432 x 1000) / 3600

286432 / 3600

7.956m/s

Therefore, the average velocity of the whole journey from beginning of the drive to the arrival at the filling station will be approximately 8 m/s

Learn more about velocity here: brainly.com/question/6504879

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6 0
3 years ago
what is the value of the constant for a second order reaction if the reactant concentration drops from .657 M to ,0981 M in 17 s
yaroslaw [1]

Answer : The value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

Explanation :

The expression used for second order kinetics is:

kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}

where,

k = rate constant = ?

t = time = 17s

[A_t] = final concentration = 0.0981 M

[A_o] = initial concentration = 0.657 M

Now put all the given values in the above expression, we get:

k\times 17s=\frac{1}{0.0981M}-\frac{1}{0.657M}

k=0.51M^{-1}s^{-1}

Therefore, the value of the constant for a second order reaction is, 0.51M^{-1}s^{-1}

6 0
3 years ago
What happens when you change the number of electrons in an atom
aalyn [17]
<h2>Answer: It becomes an Ion </h2>

When an atom has gained or lost electrons (negative charge), it becomes an ion.

In this sense:

<h2>Ions are atoms that have <u>gained or lost</u> electrons in their electronic cortex. </h2><h2> </h2>

If a neutral atom <u>loses electrons</u>, it remains with an excess of positive charge and transforms into a positive ion or <u>cation</u>, whereas if a neutral atom <u>gains electrons</u>, it acquires an excess of negative charge and transforms into a negative ion or <u>anion</u>.

It is then how ions form bonds with other atoms differently depending on the number of electrons they have.

8 0
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The technological challenges of installing cellphones across the globe
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Answer:

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3 years ago
In Challenge Example 11.9 (p. 280), after the explosion, suppose that the m1 fragment shot directly north at 12 m/s and the m3 f
Inga [223]

The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.

Solution:

It is given that mass of object before explosion is,m = 10 g

Speed of object before explosion, v = 2 m/s

Let $m_1, m_2 \text{ and}\ m_3$ be the masses of the three fragments.

Let $v_1, v_2 \text{ and}\ v_3$ be the velocities of the three fragments.

Therefore, according to the law of conservation of momentum,

$mv=m_1v_1 +m_2v_2+m_3v_3$

$10 \times 2  \hat i=3 \times 12 \hat{j} + 3(v_{2x} \hat{i}+v_{2y} \hat{j})-4 \times 9 \hat{j}$

So the x- component of the velocity of the m2 fragment after the explosion is,

$3v_{2x} = 20$

∴ $v_{2x} = 6.67 \ m/s$

6 0
3 years ago
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