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2 years ago

At which points on the roller coaster is the car not moving?

Physics

1 answer:

Scrat [10]2 years ago

3 0

The maximum potential energy of the car or the when car is not moving at points A and E; option A.

<h3>What is potential energy?</h3>

Potential energy is the energy of a body due to its state or position.

In the rollercoaster motion shown, the maximum potential energy occurs when the car is no longer moving.

At different points other than at maximum potential energy, the energy is a combination of potential and kinetic energy.

The maximum potential energy occurs at A and E, at which point the car is not moving.

In conclusion, at maximum potential energy, the car is not moving.

Learn more about potential energy at: brainly.com/question/14427111

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John ( body mass= 65 kg) is taking off for a long jump . Horizontal accerleration ax is 5m/s^2 and vertical acceleration ay is 0

cestrela7 [59]

Answer:

(a) The horizontal ground reaction force $F_{g,x}=325\, N$

(b) The vertical ground reaction force $F_{g,y}=696\, N$

(c) The resultant ground reaction force $F_g=768\, N$

Explanation:

Given

John mass , m = 65 kg

Horizontal acceleration , $a_x= 5.0 \frac{m}{s^{2}}$

Vertical acceleration , $a_y=0.9 \frac{m}{s^{2}}$

(a) Using Newton's 2nd law in horizontal direction

$F_{g,x}=ma_x$

=> $F_{g,x}=65\times 5\, N=325\, N$

Thus the horizontal ground reaction force $F_{g,x}=325\, N$

(b) Using Newton's 2nd law in vertical direction

$F_{g,y}-mg=ma_y$

=> $F_{g,y}=mg+ma_y$

=> $F_{g,y}=65\times (9.81+0.9)\, N=696\, N$

Thus the vertical ground reaction force $F_{g,y}=696\, N$

$F_g=(F_{g,x}^{2}+F_{g,y}^{2})^{\frac{1}{2}}$

=> $F_g=(325^{2}+696^{2})^{\frac{1}{2}}\, N=768\, N$

=> $F_g=768\, N$

Thus the resultant ground reaction force $F_g=768\, N$

5 0

3 years ago

Suppose that a sound has initial intensity β1 measured in decibels. This sound now increases in intensity by a factor f. What is

topjm [15]

Answer:

β2= β1+10*f

Explanation:

comparing β2 and β1, it is said that β2 is increased by a factor of f.

for each factor of f, there is a 10*f dB increase.

therefore if the β1 is increases by an intensity of factor f

the new intensity would be β1+ 10*f

4 0

3 years ago

a race car accelerates uniformly from 18.5 m/s to 46.1m/s in 2.47 seconds detrrmine the acceleration of the car and distance tra

LenaWriter [7]

Because acceleration is constant, the acceleration of the car at any time is the same as its average acceleration over the duration. So

$a=\dfrac{\Delta v}{\Delta t}=\dfrac{46.1\,\frac{\mathrm m}{\mathrm s}-18.5\,\frac{\mathrm m}{\mathrm s}}{2.47\,\mathrm s}=11.2\,\dfrac{\mathrm m}{\mathrm s^2}$

Now, we have that

${v_f}^2-{v_0}^2=2a\Delta x$

so we end up with a distance traveled of

$\left(46.1\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(18.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(11.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)\Delta x$

$\implies\Delta x=79.6\,\mathrm m$

6 0

3 years ago

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

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