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sukhopar [10]
3 years ago
9

ANSWERS UNDER QUESTIONS:

Physics
2 answers:
Nataliya [291]3 years ago
7 0

Answer:

B and E

Explanation:

Julli [10]3 years ago
6 0
<h2>Answers:</h2><h2 /><h2>a) Arrow B</h2><h2>b) Arrow E</h2>

Explanation:

Refraction is a phenomenon in which a wave (the light in this case) bends or changes its direction <u>when passing through a medium with a refractive index different from the other medium.</u>  Where the Refractive index is a number that describes how fast light propagates through a medium or material.  

According to this, if we observe the rays  A an D passing throgh the biconcave lens, we will have two mediums:

1) The air

2)The material of the biconcave lens

This two mediums have different refractive indexes, hence the rays will change the direction.

-For the incident ray A, the corresponding refractive ray is B, because is the ray that bends after passing throgh the lens

-For the incident ray D, the refracted ray is E following the same principle.

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A certain planet has a radius of 4990 km. If, on the surface of that planet, a 95.0 kg object has a weight of 591 N, then what i
aniked [119]

Answer:

3743.489 kg

Explanation:

F_g = 591 N

G = 6.674x10^-11 constant of gravity

m_1 = 95 kg

m_2 = unknown

r = 4990*1000 =

F_g = G[(m_1*m_2)/r^2]

591 N = 6.674x10^-11[(95*m_2)/4990^2]

8.855 = [(95*m_2)/4990^2]

355631.472 = 95*m_2

m_2 = 3743.489 kg

7 0
3 years ago
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)A buoy is anchored to the ocean floor. A large wave approaches the b
dimaraw [331]

Answer:

bounce up and down

Explanation:

Buoys are used for two main reasons, one is to let the people on land know of a big incoming wave, while the second reason is to generate electricity. When a big wave is approaching the buoy starts to bounce up and down with the strength of the smalled previous waves and then bounce very strongly up as the bigger wave passes by. This movement is combined with pistons within the buoy in order to conduct electricity.

8 0
3 years ago
A 9.00 V battery has a 55.0 Ohm
svetoff [14.1K]

Answer:1.47

Explanation:

I got it correct on accelus

4 0
2 years ago
1. Calcular la masa de mercurio que pasó de 30 °C hasta 120 °C y absorbió 4400 cal. Calor específico del
timofeeve [1]

Answer:

Masa, m = 0.088 kg

Explanation:

Given the following data;

Temperatura inicial = 30°C

Temperatura final = 120°C

Capacidad calorífica específica = 138J/kg.K

Calor absorbido, Q = 4400 cal.

Para encontrar la masa;

La capacidad calorífica viene dada por la fórmula;

Q = mct

Dónde;

Q representa la capacidad calorífica o la cantidad de calor.

m representa la masa de un objeto.

c representa la capacidad calorífica específica del agua.

dt representa el cambio de temperatura.

dt = T2 - T1

dt = 120 - 30

dt = 90°C to kelvin = 273 + 90 = 363K

Sustituyendo en la fórmula, tenemos;

4400 = m*138*363

4400 = 50094m

m = \frac {4400}{50094}

Masa, m = 0.088 kg

7 0
3 years ago
Matt is driving his car around a curve that has a radius of 40 m. If his speed is 25 m/s as he negotiates the curve, find the ce
s344n2d4d5 [400]

Answer:

\huge\boxed{\sf a_{c} = 15.6\ m/s^2}

Explanation:

<u>Given Data:</u>

Radius = r = 40 m

Speed = v = 25 m/s

<u>Required:</u>

Centripetal Acceleration = \sf a_{c} = ?

<u>Formula:</u>

\sf a_{c} = \frac{v^2 }{r}

<u>Solution:</u>

\sf a_{c} = \frac{v^2}{r} \\\\a_{c} = \frac{(25)^2}{40} \\\\a_{c} = \frac{625}{40} \\\\a_{c} = 15.6 m/s^2\\\\\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3>
7 0
3 years ago
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