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liberstina [14]

2 years ago

15

A net force of –8750N is used to stop of 1250.kg car travelling 25m/s. What braking distance is needed to bring the car to a hal

t?

1 answer:

Karolina [17]2 years ago

4 0

Answer:

d = 44.64 m

Explanation:

Given that,

Net force acting on the car, F = -8750 N

The mass of the car, m = 1250 kg

Initial speed of the car, u = 25 m/s

Final speed, v = 0 (it stops)

The formula for the net force is :

F = ma

a is acceleration of the car

$a=\dfrac{F}{m}\\\\a=\dfrac{-8750}{1250}\\\\a=-7\ m/s^2$

Let d be the breaking distance. It can be calculated using third equation of motion as :

$v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{0^2-(25)^2}{2\times (-7)}\\\\d=44.64\ m$

So, the required distance covered by the car is 44.64 m.

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What is the total resistance in the circuit? (include unit in answer - ohms)

solmaris [256]

Answer:60 ohms

Explanation:

R1=30 ohms

R2=15 ohms

R3=15 ohms

Let the total resistance be R

R=R1 + R2 + R3

R=30 + 15 +15

R=60

Total resistance is 60 ohms

4 0

2 years ago

Please help me with this question, especially the angle part. I have solved the initial velocity already. I don't know what to d

Helga [31]

I think you almost got it.

At the top, the velocity only has horizontal component, so v=12 m/s is v_x, which is v*cos(theta), because v_x is constant, so the same when it was launched or now.

With the value of the initial speed (28 m/s, which is the total speed), you can set

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or theta = 64.62 deg, it is D. Think about it. I hope you see it.

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2 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

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Answer:

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Since its a compound, it's charge is 0.

So, there are 2 Fe and 3 Oxygen,

let x = oxidation number of oxygen

2(+3) + 3x =0

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Side note : even if the oxidation number is positive, make sure to put the positive sign like how you would in numbers

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Free_Kalibri [48]

Answer:

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Charge on proton, $q=1.6\times 10^{-19}\ C$

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The Lorentz force is given by :

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From equation (1) and (2), we get :

$mg=qvB$

$B=\dfrac{mg}{qv}$

$B=\dfrac{1.67\times 10^{-27}\ kg\times 9.8\ m/s^2}{1.6\times 10^{-19}\ C\times 4\times 10^6\ m/s}$

$B=2.55\times 10^{-14}\ T$

Hence, this is the required solution.

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2 years ago