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stellarik [79]
2 years ago
6

Your car is initially at rest when your hit that gas and the car begins to accelerate at a rate of 1.464 m/s/s. The acceleration

lasts for 10.8 s. What is the final speed of the car and how much ground does it cover during this acceleration?​
Physics
1 answer:
tamaranim1 [39]2 years ago
6 0

Answer:

A force 1500 n to a 100 kg car . what is the acceleration of the car

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If a 100-N net force acts on a 50-kg car, what will the acceleration of the car be?

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If a net force of 5.0 N forward acts on a 0.60 kg toy car what is its acceleration

Explanation:

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max2010maxim [7]

Answer:

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4 0
2 years ago
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In order for an ion to have a -1 charge, it must have
Jet001 [13]
Gaining electron makes it “negative”
3 0
2 years ago
A crane lifts an l-bean to a height of 30 meters giving it a potential energy of 108,486). What is the mass of the I-beam?
andrew-mc [135]

369 kg

Explanation:

Step 1:

We know that the expression for potential energy is

PE = mgh

where, m =Mass

            g = Acceleration due to gravity

            h Height

Step 2:

Here m = ?

         PE = 108486

         h = 30 m

          g = 9.8 ms⁻²

108486 = 30 * 9.86 * m

m= 108486 / (9.8*30) = 369 kg

5 0
2 years ago
When milk goes sour, what type of change has happened?
zzz [600]
B) Chemical change only
7 0
3 years ago
Read 2 more answers
You have a 1 W light bulb in your lab. It puts out light of only 1 frequency. The wavelength of this light is 500nm. you set up
ollegr [7]

Answer:

a)   # _photon = 2.5 10¹⁸ photons / s,   b) E = 10⁻² N / C,  c)     B = 3 10⁻¹¹ T

d)  r=  2 10⁹ m

Explanation:

a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation

          E₀ = h f

          c = λ f

          E₀ = h c /λ

          E₀ = 6.63 10⁻³³⁴   3 10⁸/500 10⁻⁹

          E₀ = 3.978 10⁻⁻¹⁹ J

Let's use a direct ratio rule to find the number of photons

         #_foton = E / Eo

         #_fototn = 1 / 3.978 10⁻¹⁹

         # _photon = 2.5 10¹⁸ photons / s

b) The intensity received by the detector is related to the electric field

          I = E²

Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space

          I = P / A

          P = I A

Let's use index 1 for the point on the bulb and index 2 for the point on the detector.

The area of ​​a sphere is

          A = 4π r²

         P = I₁ A₁ = I₂ A₂

         I₁ r₁² = I₂ r₂²

         I₂ = I₁  r₁²/r₂²

         I₂ = I₁    1 / 100²

         I₂ = I₁ 10⁻⁴

we must know the intensity at the output of the bulb suppose that I₁ = 1 J

          I₂ = 10⁻⁴ J

let's look for the electric field

         E =√I

         E = √10⁻⁴

         E = 10⁻² N / C

c) for the calculation of the magnetic field we use that the field is in phase

               E / B = c

               B = E / c

               B = 10⁻² / 3 10⁸

               B = 3 10⁻¹¹ T

d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area  A = 4π R² where R = 100 m how many photons are there in the area of ​​the detector r = 1 cm,   A’= 10⁻⁴ m²

             #_photons = 2.5 10¹⁸ A_detector / A_sphere

             #_photons = 2.5 1018 10-4 / 4π 10⁴

             #_photons = 2 10⁹ photons in the detector area

for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m

6 0
2 years ago
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