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s344n2d4d5 [400]
3 years ago
12

Please follow I will follow you​

Physics
2 answers:
svet-max [94.6K]3 years ago
7 0
Huh wdym, follow what,,,
klio [65]3 years ago
5 0

Answer:

aight

Explanation:

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An object is placed at 10.2 cm in front of a diverging lens with a focal length of -10.6 cm. What is the magnification
adoni [48]

Answer:

M= -0.51

Explanation:

After i calculated my v to be -5.2cm from the formula 1/f=1/v+1/u

Then m=v/u which is -0.51

7 0
4 years ago
A brick of mass 2.0kg is at rest. It falls to the ground through a
Lisa [10]

Answer:

I may not have the answer so i'll just give up some hints.

Multiply the time by the acceleration due to gravity to find the velocity when the object hits the ground. If it takes 9.9 seconds for the object to hit the ground, its velocity is (1.01 s)*(9.8 m/s^2), or 9.9 m/s. Choose how long the object is falling. In this example, we will use the time of 8 seconds. Calculate the final free fall speed (just before hitting the ground) with the formula v = v₀ + gt = 0 + 9.80665 * 8 = 78.45 m/s . Find the free fall distance using the equation s = (1/2)gt² = 0.5 * 9.80665 * 8² = 313.8 m .h = 0.5 * 9.8 * (1.5)^2 = 11m. b. V = gt = 9.8 * 1.5 = 14.7m/s. A feather and brick dropped together. Air resistance causes the feather to fall more slowly. If a feather and a brick were dropped together in a vacuum—that is, an area from which all air has been removed—they would fall at the same rate, and hit the ground at the same time.When an object's point is taller the thing that is going down it will go faster than when the point is lower. EXAMPLE: The object is the tennis ball if you drop it down the higher hill it will be faster than if you drop it down a shorter hill. In other words, if two objects are the same size but one is heavier, the heavier one has greater density than the lighter object. Therefore, when both objects are dropped from the same height and at the same time, the heavier object should hit the ground before the lighter one.

I hope my little bit (big you may say) hint help you with your question.

5 0
3 years ago
Brainliest for whoever is first and right.
devlian [24]

The answer is 9.8, did this last year in AP Science

3 0
3 years ago
WILL UPVOTE! MULTIPLE CHOICE!
Stels [109]
1. A. The SI unit for current is the ampere.
2. C. The unit for resistance = ohm.
3. B. EMF is measured in volts.
4. B. Electric current is the flow of charged particles, i.e. electrons.
5. B or C. All other answers are metals.
6. C. The dimensions of a material don't affect its resistance.
7. D. Ohm's law: current = voltage / resistance, 12/4 = 3.
8. B. Batteries transform chemical energy into electrical energy.
9. A. Metals are generally good conductors.
6 0
4 years ago
An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose th
igomit [66]

Answer:

T = 1.12 10⁻⁷ s

Explanation:

This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.

All the charge dq is at a distance r

           dE = k dq / r²

Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry

            cos θ =\frac{dE_x}{dE}

             dEₓ = dE cos θ

              cos θ = x / r

substituting

                dEₓ = k \frac{dq}{r^2 } \ \frac{x}{r}

                DEₓ = k dq x / r³

let's use the Pythagorean theorem to find the distance r

             r² = x² + a²

where a is the radius of the ring

we substitute

              dEₓ = k \frac{x}{(x^2 + a^2 ) ^{3/2} } \ dq

we integrate

               ∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} }  ∫ dq

               Eₓ = k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}

In the exercise indicate that the electron is very central to the center of the ring

                x << a

                Eₓ = k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}

if we expand in a series

                  (\ 1+ (x/a)^2 \  )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2

we keep the first term if x<<a

                 Eₓ = \frac{ k Q}{a^3} \ x

the force is

                 F = q E

                 F = - \frac{kQ  }{a^3} \ x

this is a restoring force proportional to the displacement so the movement is simple harmonic,

                 F = m a

                 - \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }

                 \frac{d^2 x}{dt^2} = \frac{keQ}{ma^3}  \ x

the solution is of type

                  x = A cos (wt + Ф)

with angular velocity

                w² = \frac{keQ}{m a^3}

angular velocity and period are related

                 w = 2π/ T

             

we substitute

               4π² / T² = \frac{keQ}{m a^3}

                T = 2π  \sqrt{\frac{m a^3 }{keQ} }

let's calculate

                T = 2π \sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19}  \ 0.021  \ 10^{-3} }  }

                 T = 2π pi \sqrt{320.426 \ 10^{-18} }

                 T = 2π  17.9 10⁻⁹ s

                 T = 1.12 10⁻⁷ s

6 0
3 years ago
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