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algol [13]
3 years ago
11

At the bottom of a loop in the vertical plane, an airplane has a horizontal velocity of 375 mi/h and is speeding up at a rate of

10 ft/s. the radius of curvature of the loop is 1 mi. the plane is being tracked by radar at o. what are the recorded values of formula773.mml for this instant?

Physics
1 answer:
Sedaia [141]3 years ago
5 0
Check the attached file for the answer.

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What is the potential difference when the current in a circuit is 5mA and resistance is 30 Ohms
Mashcka [7]

<h2>\bf{ \underline{Given:- }}</h2>

\sf• \: The \:  current \:  in \:  a \:  circuit \:  is  \: 5 \: amps.  \: and  \: resistance \:  is \:  30 \:  Ohms.

\\

<h2>\bf{ \underline{To \:  Find :- }}</h2>

\sf{• \:  The  \: Potential  \: Difference. }

\\

\huge\bf{ \underline{ Solution:- }}

\sf According  \: to  \: the  \: question,

\sf•  \: Current \:  (I) = 5  \: Amps.

\sf• \:  Resistance  \: (R) = 30 \:  Ω

\sf{Potential \:  difference  \: means  \: Voltage \: ( V).}

\sf{We \:  know \: that, }

\bf \red{ \bigstar{ \: V = IR }}

\rightarrow \sf V =5 \times 30

\rightarrow \sf V =150

\\

\sf \purple{Therefore, \:  the \:  potential  \: difference  \: is  \: 150  \: v \: .}

4 0
3 years ago
Nuclear fusion is when two atoms of __________________ join together to form _____________.
solong [7]

Answer:

1. Hydrogen
2. Helium

Explanation:

Nuclear fusion is when two atoms of Hydrogen join together to form one Helium atom.

3 0
2 years ago
an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati
kicyunya [14]
GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
I hope this helps ~~Charlotte~~
5 0
2 years ago
Read 2 more answers
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

6 0
3 years ago
2. A construction consultant may be responsible for
Brrunno [24]

This is a tricky one but on my part I'd have to say depending on the contract A,B,C.

3 0
3 years ago
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