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3 years ago

Due today HELP HELP HELP

Physics

1 answer:

Anna [14]3 years ago

5 0

Vas happenin!

Independent variable : amount of water each day

Dependent variable: water on the windsill

Hypotheses: Ben wants to try by adding water each day to two different places. Will that work? Will that effect the water?

Hope this helps you out

*smiles*

-Zayn Malik

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I am trying to brush up on my Physics skills and I came across a couple of equations that I cannot seem to remember what they ar

Aleks04 [339]

Explanation:

Eq 2 ) is Equation for force on a charge Q1 due to charge Q2 separated by a distance R .. it is Equation for Coloumb's law !!

Where , k = 1/4π€

Eq 3 ) is Equation for force on mass m1 due to mass m2 separated by distance R .. it is Newton's law of Gravitation

Where , G = universal Gravitation constant

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3 years ago

Newton's second law of motion states the relationship of mass, acceleration and force. It states that

sergey [27]

Answer:

Explanation:

force equals mass time acceleration

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3 years ago

T=2pi square root 1/g solve for g.<br> Explanation would be really helpful.

Natalija [7]

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

$T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}$

Let me know if you have any questions.

7 0

3 years ago

A world-class sprinter can reach a top speed (of about 11.8 m/s ) in the first 19.0 m of a race. what is the average acceleratio

Whitepunk [10]

We use the kinematic equation of motion, to calculate the average acceleration

$v^{2} =u^{2} +2a (h-h_{0} )$

or $a=\frac{v^{2} -v_{0}^2}{2(h-h_{0}) }$

Here, $v_{0}$ is initial speed , v is final speed h is covered distance $h_{0}$ is initial position.

Given $v= 11.8 m/s$ , $v_{0} =0$ , $h = 19.0 m$ and $h_{0} =0$ .

Substituting these values in above equation we get,

$a=\frac{(11.8)^2-0}{2(19.0-0)} \\\\\ a =3.66 m/s^2$

Thus, the average acceleration of the sprinter is $3.66 m/s^2$ .

6 0

3 years ago

A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−

Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ .

Explanation:

Given that,

The actual wavelength of the hydrogen atom, $\lambda_a=5.13\times 10^{-7}\ m$

A hydrogen atom in a galaxy moving with a speed of, $v=6.65\times 10^6\ m/s$

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

$v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}$

$\lambda_o$ is the observed wavelength

$\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m$

So, the observed wavelength on Earth from that hydrogen atom is $5.24\times 10^{-7}\ m$ . Hence, this is the required solution.

8 0

3 years ago