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larisa [96]
3 years ago
6

Answer the question fast please​

Physics
1 answer:
AnnZ [28]3 years ago
6 0

Answer:

1

Explanation:

d = m \div v  \\ m = 0.8 \: v = 80\% \: also \: 0.8 \\ d = 0.8 \div 0.8 \\ d = 1

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Serggg [28]
ANY orbiting object travels fastest when it's closest to the central body.
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Which of the following is true of an adiabatic process?
poizon [28]

Adiabatic process is that process in which heat of a closed system will remain conserved or we can say that system can not exchange its thermal energy from its surrounding

So here from 1st law of thermodynamics we can say

\Delta Q = \Delta U + W

since we know that

\Delta Q = 0

so we will have

0 = \Delta U + W

so we have

-\Delta U = W

so here correct answers are

<em>A. -ΔU = W  </em>

<em>B. Heat is constant. </em>

7 0
4 years ago
Read 2 more answers
An object moves 2.5 m. This is an example of a _______.
makvit [3.9K]

The correct answer is A. Distance

Hope this helps

7 0
3 years ago
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If the third charge (–| q 3 |) is placed at point P, but not held fixed, it will experience a force and accelerate away from the
Mazyrski [523]

Complete Question

The complete question is shown on the first uploaded image

Answer:

A

The potential of this system is  U=6.75*10^{-7}J

B

The electric potential at point p is V_p= -900V

C

The work required is  W= 9*10^{-7}J

D

The speed of the charge is  v=600m/s

Explanation:

A sketch to explain the question is shown on the second uploaded image

Generally the potential energy for a system of two charges is mathematically represented as

            U = \frac{kq_1 q_2}{d}

where k is the electrostatic constant with a value of  k = 9*10^9 N m^2 /C^2

           q is the charge with a value of  q = 1*10^{-9}C

           d is the distance given as   d =5m

Now we are given that  q_1 = q and  q_2 = 3q and

Now substituting values

             U = \frac{9*10^9 *1*10^{-9} * 3*10^{-9}}{5}

                U=6.75*10^{-7}J

The electric potential at point P is mathematically obtained with the formula

             V_p = V_{-q} + V_{-3q}

I.e the potential at q_1 plus the potential at  q_2

Now potential at q_1 is mathematically represented as

                   V_{-q} = \frac{-kq}{s}

and the potential at q_2 is mathematically represented as

                        V_{-3q} = \frac{-3kq}{s}

Now substituting into formula for potential at  P

                  V_p = \frac{-kq}{s} + \frac{-3kq}{s}  = -\frac{4kq}{s}

                       = \frac{4*9*10^9 *1*10^{-9}}{4*10^{-2}}

                      V_p= -900V

The Workdone to bring the third negative charge is mathematically evaluated as

                W =\Delta U = \frac{kq_1q_3}{s} + \frac{kq_2q_3}{s}

                                 = \frac{kq*q}{s} +   \frac{kq*3q}{s}

                                = \frac{4kq^2}{s}

                               = \frac{4* 9*10^9 * (1*10^{-9})^2}{4*10^{-2}}

                              W= 9*10^{-7}J

From the Question are told that the charge q_3 would a force and an acceleration which implies that all its potential energy would be converted to kinetic energy.This can be mathematically  represented as

                   \Delta U = W = \frac{1}{2} m_{q_3} v^2

                         9*10^{-7} = \frac{1}{2} m_{q_3} v^2

Where m_{q_3} = 5.0*10^{-12}kg

Now making v the subject we have

                 v = \sqrt{\frac{9*10^{-12}}{5*10^{-12}*0.5} }

                     v=600m/s  

8 0
4 years ago
I NEED HELP PLEASE, THANKS! Light passes from air into water at an angle of 40.0° to the normal. What is the angle of refraction
Vlad [161]

Answer:

Angle of Refraction = 28.9 degrees

Explanation:

<u><em>We'll use Snell's law for this. It's mathematical form is:</em></u>

=> n_1* sin(\alpha _1)=n_2 * sin(\alpha _2)

Where n_{1} = 1, n_{2} = 1.33, \alpha _{2} = 40^o

=> n_{1} and n_{2} are the refractive indexes of the air and water respectively.

<u><em>Solution:</em></u>

=> 1 * sin (40) = 1.33 * sin(\alpha _{2})

=> sin \alpha _{2} = \frac{sin 40^0}{1.33}

=> sin \alpha _{2} = 0.4821

=> \alpha _{2} = 28.9 degrees

8 0
3 years ago
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