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3 years ago

A car accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car’s acceleration is __meters/second2.

Physics

2 answers:

Andrej [43]3 years ago

4 0

3 meters.............

Mariana [72]3 years ago

3 0

3 meters/second2
a=v-u/t

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The most crucial information would be its atomic number.

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3 years ago

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3 years ago

A draft is made of a plastic block with a density of 650kg/m^3, and its dimensions are 2.0mx3.0mx5.0m ,what is the raft's appare

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3 years ago

If f(x)=4/x+2 and g is the inverse of f,then g'(10)=

ch4aika [34]

Answer:

g'(10) = $\frac{-1}{16}$

Explanation:

Since g is the inverse of f ,

We can write

g(f(x)) = x <em> </em><em>(Identity)</em>

Differentiating both sides of the equation we get,

g'(f(x)).f'(x) = 1

g'(10) = $\frac{1}{f'(x)}$ --equation[1] Where f(x) = 10

Now, we have to find x when f(x) = 10

Thus 10 = $\frac{4}{x}$ + 2

$\frac{4}{x}$ = 8

x = $\frac{1}{2}$

Since f(x) = $\frac{4}{x}$ + 2

f'(x) = - $\frac{4}{x^{2} }$

f'( $\frac{1}{2}$ ) = -4 × 4 = -16

Putting it in equation 1, we get:

We get g'(10) = - $\frac{1}{16}$

5 0

3 years ago

An electron in the n = 7 level of the hydrogen atom relaxes to a lower energy level, emitting light of 397 nm. what is the value

Dimas [21]

Answer:

$n_f=2$

Explanation:

It is given that,

Initially, the electron is in n = 7 energy level. When it relaxes to a lower energy level, emitting light of 397 nm. We need to find the value of n for the level to which the electron relaxed. It can be calculate using the formula as :

$\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})$

$\dfrac{1}{397\times 10^{-9}\ m}=R(\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

R = Rydberg constant, $R=1.097\times 10^7\ m^{-1}$

$\dfrac{1}{397\times 10^{-9}\ m}=1.097\times 10^7\ m^{-1}\times (\dfrac{1}{n_f^2}-\dfrac{1}{(7)^2})$

Solving above equation we get the value of final n is,

$n_f=2.04$

$n_f=2$

So, it will relax in the n = 2. Hence, this is the required solution.

6 0

3 years ago

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