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3 years ago

A girl on the outer edge of a merry-go-round moves in uniform circular motion. Which of the following is NOT correct?

Physics

2 answers:

nignag [31]3 years ago

7 0

I think that from the answers above the answer is B.

hichkok12 [17]3 years ago

3 0

This is a poor question. I think they want you to choose D because the girl is traveling in a circular path. But a circle is an ellipse. So technically D is also true.

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2 years ago

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3 years ago

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Find the Y component of 35m/s at 57 degrees from the X-axis.

rosijanka [135]

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3 years ago

A driver in a car traveling at a speed of 50 mi/h sees a deer 50 m away on the road. Calculate the minimum constant acceleration

alisha [4.7K]

Answer:

a= - 0.79 m/s²

Explanation:

Given that

Speed ,u = 20 mi/h

We know that

1 mi/h= 0.44 m/s

Therefore ,u = 8.94 m/s

Distance ,s= 50 m

Lets take the acceleration of the car = a m/s²

The final speed of the car ,v = 0 m/s

We know that

v²= u² + 2 a s

Now by putting the values

0²= 8.94² + 2 x a x 50

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a= - 0.79 m/s²

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6 0

3 years ago

The Ha line of the Balmer series is emitted in the transition from n-3 to n 2. Compute the wavelength of this line for H and 2H.

Citrus2011 [14]

Explanation:

According to Rydberg's formula, the wavelength of the balmer series is given by:

$\frac{1}{\lambda}=R(\frac{1}{2^2}-\frac{1}{3^2})$

R is Rydberg constant for an especific hydrogen-like atom, we may calculate R for hydrogen and deuterium atoms from:

$R=\frac{R_{\infty}}{(1+\frac{m_e}{M})}$

Here, $R_{\infty}$ is the "general" Rydberg constant, $m_e$ is electron's mass and M is the mass of the atom nucleus

For hydrogen, we have, $M=1.67*10^{-27}kg$ :

$R_H=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{1.67*10^{-27}kg})}\\R_H=1.09677*10^7m^{-1}$

Now, we calculate the wavelength for hydrogen:

$\frac{1}{\lambda}=R_H(\frac{1}{2^2}-\frac{1}{3^2})\\\lambda=[R_H(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.0967*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5646*10^{-7}m=656.46nm$

For deuterium, we have $M=2(1.67*10^{-27}kg)$ :

$R_D=\frac{1.09737*10^7m^{-1}}{(1+\frac{9.11*10^{-31}kg}{2*1.67*10^{-27}kg})}\\R_D=1.09707*10^7m^{-1}\\\\\lambda=[R_D(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=[1.09707*10^7m^{-1}(\frac{1}{2^2}-\frac{1}{3^2})]^{-1}\\\lambda=6.5629*10^{-7}=656.29nm$

5 0

3 years ago