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Anni [7]
3 years ago
7

A car moving south speeds up from 10 m/s to 40 m/s in 15 seconds. What is the car’s acceleration?

Chemistry
2 answers:
faltersainse [42]3 years ago
6 0
Answer is 2m/s sq because V-u /t us acceleration 40-10/15=2
Firlakuza [10]3 years ago
6 0

Answer : The car’s acceleration is, 2m/s^2

Solution :

Acceleration : It is defined as the change in the velocity of an object with respect to the time.

Formula used :

a=\frac{v_{final}-v_{initial}}{t}

where,

a = acceleration of car = ?

v_{final} = final velocity of car = 40 m/s

v_{initial} = initial velocity of car = 10 m/s

t = time taken = 15 s

Now put all the given values in this formula, we get the acceleration of car.

a=\frac{40m/s-10m/s}{15s}=2m/s^2

Therefore, the acceleration of car is, 2m/s^2

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maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


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Did diagram shows embryo development of four different animals. How is this evidence used to suggest that life changes over time
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Answer:

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The n and l values and the number of orbitals for sublevel 5g is :

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1) Principal quantum number , n

2) Angular quantum number , l

3) Magnetic quantum number , ml

4) spin quantum number , ms

For 5g shell, n = 5

subshell g , l = 4     ....0 - s , 1 - p , 2 - d, 3 - f, 4 -g

number of orbitals in subshell = (2l + 1)  ( 2×4 + 1) = 9

Thus,  The n and l values and the number of orbitals for sublevel 5g is :

5g shell , n= 5 subshell g , l = 4, Number of orbitals for sublevel = 9.

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