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disa [49]
3 years ago
10

What average force is required to stop a 980 kg car in 9.0 s if the car is traveling at 91 km/h ?

Physics
1 answer:
kozerog [31]3 years ago
5 0
Newton's second law of motion:
F=ma
a =  \frac{v}{t}
You convert from km/h to m/s by dividing by 3.6:
v = 91 \div 3.6
v = 25.3 \:  \frac{m}{s}
Then a is:
a =  \frac{25.3}{9}
a = 2.8  \frac{m}{s {}^{2} }
Then:
F=(980)(2.8)=2744 N
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The intensity of electromagnetic radiation from the sun reaching the earth's upper atmosphere is 1.37kW/m2kW/m2. Part A Assuming
Juli2301 [7.4K]

Answer:

#_photon = 7  10²¹ photons

Explanation:

Let's look for the power that affects the panel of area of ​​1.5 m2

           I = P / A

           P = I A

           P = 1.37 10³  1.5

           P = 2,055 10³ W

           P = E / t

       

If we use t = 1 s

           E = P t

           E = 2,055 10³ J

This is the power that the panel receives, let's look for the energy of a photon

            E = h f

            c = λ f

            f = c /λ

            E = h c /λ

Let's calculate

            E₀ = 6.63 10⁻³⁴  3 10⁸/680 10⁻⁹

            E₀ = 2.925 10⁻¹⁹ J

In one second the total energy is the number of photons for the energy of each one

             E = #_photon  E₀

             #_photon = E / E₀

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8 0
3 years ago
An object is 40 cm in front of a convex lens with a focal length of 20 cm. What is the position of the image using ray tracing?
mr Goodwill [35]

Answer:

40 cm behind the lens

Explanation:

Please refer to the attached picture for the ray tracing, where

f is the focal length

p is the location of the object

q is the location of the image

Each tick in the picture corresponds to 20 cm, so we can observe that

p = 40 cm

f = 20 cm

q = 40 cm

Also, q is inverted, real and same size as the object.

The location of the image, q, can also be verified by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{20 cm}-\frac{1}{40 cm}=\frac{1}{40 cm}\\q = 40 cm

3 0
3 years ago
Gravity and magnetism are _ forces.<br>A. contact<br>B. field<br>C. tension<br>D. normal​
e-lub [12.9K]

Answer:

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A normal force is one that is perpendicular to the direction of a slope, so that eliminates option D.

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