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Andrew [12]
3 years ago
9

In which of the following situation is light most likely to be a reflection

Physics
1 answer:
maxonik [38]3 years ago
7 0

Answer:

when the reflecting surface is plain and without even small hurdles that are not the visible by our naked eyes. Eg : plain mirror

Explanation:

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The horse on a carousel is 3.5m from the central axis.A. If the carousel rotates at 0.13 rev/s , how long does it take the horse
Kitty [74]

Answer:

a. 15.4 seconds

b. 0.455 m/s

Explanation:

a. The carousel rotates at 0.13 rev/s.

This means that it takes the carousel 1 sec to make 0.13 of an entire revolution.

This means that time it will take to make a complete revolution is:

1 / 0.13 = 7.7 seconds

Therefore, the time it will take to make 2 revolutions is:

2 * 7.7 = 15.4 seconds

b. Let us calculate the linear velocity. Angular velocity is given as:

\omega = v / r

where v = linear velocity and r = radius

The radius of the circle is 3.5 m and the angular velocity is 0.13 rev/s, therefore:

0.13 = v / 3.5

v = 3.5 * 0.13 = 0.455 m/s

Linear velocity is 0.455 m/s

8 0
3 years ago
BRAINLEST TO SOMEONE WHO ACTUALLY ANSWERS AND FREE 100 POINTS TO REAL ANSWER 2. Using the average time, calculate the oscillatio
andrezito [222]

Answer:

IM SO SORRY :((

Explanation:

7 0
3 years ago
Read 2 more answers
The mass of Jupiter is 1.9 × 1027 and that of the sun is 1.99 × 1030. The
n200080 [17]

Answer:

F = 4.147 × 10^23

v = 1.31 × 10^4

Explanation:

Given the following :

mass of Jupiter (m1) = 1.9 × 10^27

Mass of sun (m2) = 1.99 × 10^30

Distance between sun and jupiter (r) = 7.8 × 10^11m

Gravitational force (F) :

(Gm1m2) / r^2

Where ; G = 6.673×10^-11 ( Gravitational constant)

F = [(6.673×10^-11) × (1.9 × 10^27) × (1.99 × 10^30)] / (7.8 × 10^11)^2

F = [25.231 × 10^(-11+27+30)] / (60.84 × 10^22)

F = (25.231 × 10^46) / (60.84 × 10^22)

F = 3.235 × 10^(46 - 22)

F = 0.4147 × 10^24

F = 4.147 × 10^23

Speed of Jupiter (v) :

v = √(Fr) / m1

v = √[(4.147 × 10^23) × (7.8 × 10^11) / (1.9 × 10^27)

v = √32.3466 × 10^(23+11) / 1.9 × 10^27

v = √32.3466× 10^34 / 1.9 × 10^27

v = √17. 023 × 10^34-27

v = √17.023 × 10^7

v = 13047.221

v = 1.31 × 10^4

4 0
3 years ago
12.51 A parallel RLC circuit, which is driven by a variable frequency 2-A current source, has the following values: R = 1 kΩ, L
Anastaziya [24]

Answer:

BW = 100 rad/s

wlow = 452.49 rad/s

whigh = 552.49 rad/s

V(jwlow) =1414.21 < 45°V

V(jwhigh) =1414.21 <-45°V

Explanation:

To calculate bandwidth we have formula

BW = 1/RC

BW = 1/ 1000x10x10^¯6

BW = 100 rad/s

We will first calculate resonant frequency and quality factor for half power frequencies.

For resonant frequency

wo = 1/(SQRT LC)

wo = 1/SQRT 400×10¯³ × 10×10^¯6

wo = 500 rad/s

For Quality

Q = wo / BW

Q = 500/100

Q = 5

wlow = wo [-1/2Q+ SQRT (1/2Q)² + 1]

wlow = 500 [-1/2×5 + SQRT (1/2×5)² + 1]

wlow = 452.49 rad/s

whigh = wo [1/2Q+ SQRT (1/2Q)² + 1]

whigh = 500 [1/2×5 + SQRT (1/2×5)² + 1]

whigh = 552.49 rad/s

We will start with admittance at lower half power frequency

Y(jwlow) = (1/R) + (1/jwlow L) + (jwlow C)

Y(jwlow) = (1/1000) + (1/j×452.49×400×10¯³) + (j×452.49×10×10^¯6)

Y(jwlow) = 0.001 - j5.525×10¯³ + j4.525×10¯³

Y(jwlow) = (1-j).10¯³ S

Voltage across the network is calculated by ohm's law

V(jwlow) = I/Y(jwlow)

V(jwlow) = 2/(1-j).10¯³

V(jwlow) = 1414.2 < 45°V

Now we will calculate the admittance at higher half power frequency

Y(jwhigh) = (1/R) + (1/jwhigh L) + (jwhigh C)

Y(jwhigh) = (1/1000) + (1/j×552.49×400×10¯³) + (j×552.49×10×10^¯6)

Y(jwhigh) = 0.001 - j4.525×10¯³ + j5.525×10¯³

Y(jwhigh) = (1+j).10¯³ S

Voltage across network will be calculated by ohm's law

V(jwhigh) = I/Y(jwhigh)

V(jwhigh) = 2/(1+j).10¯³

V(jwhigh) = 1414.2 < - 45°V

6 0
3 years ago
Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva??
Brums [2.3K]
<h3>Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva?</h3>

  • Es normal que los quemadores superiores de una estufa eléctrica o una estufa se enciendan y apaguen en configuraciones distintas a Hi. El quemador se encenderá y apagará más de lo normal cuando se utilizan cacerolas que no son planas o que son del tamaño incorrecto para el quemador.

<h2>☆彡Hanna</h2>

#CarryOnLearning

7 0
3 years ago
Read 2 more answers
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