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3 years ago

A runner accelerates to 4.2 m/s2 for 10 seconds before winning the race. How far did he/she run?

Physics

2 answers:

timofeeve [1]3 years ago

8 0

She run the distance of 42metres

hodyreva [135]3 years ago

6 0

Answer:

Runner will run for 210m

Step by Step Explanation:

According to Newton's first of motion,

Vf=Vi+at

Given is

the speed with which runner is running will be Vi, he is running with accelertion a=4.2m/s2 fot time t=10sec

Vf will be 0 because runner will stop after winning the race

we will find Vi from equation 1,

Vi=Vf-at

= 0- 4.2*10

=-42 m/s (negative sign is showing decceleration)

So, Vi=42 m/s

According to Newton's third of motion,

2aS= Vf^2-Vi^2

S is distance, he will cover before stopping,

2(4.2)S= 0- 42^2

S=1764/8.4

S=210m

Runner will run for 210m

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A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm

andrew11 [14]

Answer:

2.8 cm

Explanation:

$y_1$ = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

$\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm$

Fringe width is also given by

$\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}$

For second order

$\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1$

Distance between two second order minima is given by

$y_2=2\beta_2$

$\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm$

The distance between the two second order minima is 2.8 cm

8 0

3 years ago

1. El puente mas largo del mundo es el puente Akashi Kaikyo, en Japón. El puente mide 3910 m de largo y está construido con acer

bearhunter [10]

Answer:

It's 1.0000042 times longer in summer than in winter. It represents a 1.6 centimeters difference between seasons.

Explanation:

The linear coefficient of thermal expansion for steel is about $1.2*10^{-7}\°C^{-1}$ . From the equation of linear thermal expansion, we have:

$L_f=L_0(1+\alpha\Delta T)$

Taking the winter day as the initial, and the summer day as the final, we can take the relationship between them:

$L_{summer}=L_{winter}[1+(1.2*10^{-7}\°C^{-1})(30\°C+5\°C)]\\\\L_{summer}=(1.0000042)L_{winter}$

It means that the bridge is 1.0000042 times longer in summer than in winter. If we multiply it by the length of the bridge, we obtain that the difference is of about 1.6 centimeters between the two seasons.

8 0

4 years ago

Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley

kenny6666 [7]

Answer:

Hello your question is incomplete attached below is the complete question

Answer : x ( acceleration of mass 4m ) = $\frac{g}{7}$