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IceJOKER [234]
2 years ago
11

What happens when light rays pass through a converging lens (convex lens)?

Physics
1 answer:
Hunter-Best [27]2 years ago
8 0
The light rays refract. 
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An object with a mass of 7.60 kg is moving to the right and experiences an applied force of 50 N to the right. The friction forc
horrorfan [7]

Answer:

Explanation:

We will use the equation F - f = ma, which is a fancy way of stating Newton's 2nd Law.

F = +50.0,

f = -30.0,

m = 7.60 kg. Therefore:

50.0 - 30.0 = 7.60a and

20.0 = 7.60a and

a = 20.0/7.60 so

a = 2.63 m/s/s to the right

4 0
2 years ago
A 55 kg roller skater is at rest on a flat skating rink, a 198 N horizontal force is needed to set the skater in motion.
Rufina [12.5K]

Answer:

Explanation:

To get the person Moving you have to overcome the static (means not moving) friction coefficient.  U(static)

To get the person going at the same speed you have to overcome the kinetic friction coefficient. U(Kinetic)

Force to get him moving is 198 N.   Force = ma = U(static)Mg

combining the 2 equations you get 198N = U(static)* 55kg *9.8m/s^2   Solve for U(static)

Same equation to keep him moving except with the dynamic force and the dynamic U

 

175N=  U(kinetic)*55kg*9.8m/s^2  Solve (U dynamic)

8 0
2 years ago
How much work is done on 10.0 c of charge to move it through a potential diffrence of 9.0 v in 10.0s
GenaCL600 [577]
We need more evidence to be provided
4 0
2 years ago
Zad. 2 W marszobiegu chłopiec przebiegł 900 m w ciągu 205 min, a następnie przeszedł 300m w tym danym czasie. Jaka była średnia
USPshnik [31]

Odpowiedź:

0,049 m / s

Wyjaśnienie:

Biorąc pod uwagę, że:

Dystans biegu = 900m

Czas trwania = 205 minut

Długość przejścia = 300 m

Zajęty czas = 205 minut

Średnia prędkość :

(Przebieg + pokonany dystans) / całkowity czas

Średnia prędkość :

(900 m +. 300 m) / 205 + 205

1200 m / 410 minut

Minuty do sekund

1200 / (410 * 60)

1200/24600

= 0,0487804

= 0,049 m / s

8 0
2 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
Art [367]

Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

a) satellite's orbital velocity

   centripetal force acting on satellite = \dfrac{mv^2}{r}

     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

      v = \sqrt{\dfrac{6.67 \times 10^{-11}\times 5.972 \times 10^{24}}{2 \times 6.4 \times 10^6}}

          v = 5578.5 m/s

b) T= \dfrac{2\pi\ r}{v}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

          T = 14416.92 s

          T = \dfrac{14416.92}{3600}\ hr

          T = 4 hr

c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

4 0
3 years ago
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