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aleksklad [387]
3 years ago
7

A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. It takes him 260 s to m

ake this trip.what is his speed. express answer in 2 significant figures
Physics
1 answer:
Ber [7]3 years ago
3 0

Answer:

0.7 m/s

Explanation:

The following data were obtained from the question:

Distance travalled (d) = 190 m

Time (t) = 260 secs

Speed (S) =..?

Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:

Speed = Distance (d) /time(t)

S = d/t

With the above formula, we can calculate the speed of the security guard as follow:

Distance travalled (d) = 190 m

Time (t) = 260 secs

Speed (S) =..?

S = d/t

S = 190/260

S = 0.7 m/s

Therefore, the speed of the security guard is 0.7 m/s

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If for a given pair of media CR
nika2105 [10]

CR​ < CY​ < CB​

<h3>Which factors affect the critical angle for a given pair of media?</h3>

The factors which affect the critical angle are

(a) The colour (or wavelength) of light

(b) The temperature

(i) Effect of colour of light: The critical angle for a pair of media is less for the violet light and more for the red light. Thus the critical angle increases with the increase in wavelength of light.

(ii) Effect of temperature: The critical angle increases with increase in temperature because on increasing temperature of medium, its refractive index decreases.

According to the question,

μ 1​ sinCR​ =1

μ 2​ sinCY =1

μ 3​ sinCB​ =1

μ 1​ > μ 2​  and μ 2​ > μ 3

⟹μ 1​ > μ 2​ > μ 3

CR​ < CY​ < CB​

Thus,

The critical angle increases with the increase in wavelength of light.

Learn more about wavelength of light here:

brainly.com/question/27557868

#SPJ1

7 0
2 years ago
If, in general, r were calculated as r =v/i, which circuit arrangement in part a of the experiment would have the smallest error
ladessa [460]

Concept: According to Ohm's Law, the flow of electric current through a conductor is directly proportional to the potential difference across it, provided physical conditions (like temperature, pressure, volume etc.) remains same.

v = ir

or, r = v / i

Here, current (i) is measured by Ammeter which should be connected in series of any electrical circuit.

voltage (v) is measured by Voltmeter which should be connected parallel to the external resistance (r).

In the given experiment, the first arrangement of the circuit will show the smallest error because the voltmeter is connected exactly parallel to the external resistance.

In the second arrangement, the voltmeter is connected across external resistance (r) and Ammeter (A) and in this case, the voltmeter will not measure the exact potential drop across the external resistance (r). So, there would be more error.

4 0
3 years ago
Light from a laser (lambda= 406.192 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen
dsp73

Answer:

The spacing between the slits is    d = 0.00145m                

Explanation:

From the question we are told that

  The wavelength of the light is \lambda = 406.192nm = 406.192*10^{-9} m

   The distance of the slit from the screen is D = 5.937 \ m

    The number of bright fringe is n = 24

     The  length the fringes span is   L = 39.835 mm = \frac{39.835 }{1000} = 0.0398 m

The fringe width (i.e the distance of between two successive bright or dark fringe) is mathematically represented as

             \beta  = \frac{\lambda D}{d}

Where d is  the distance between the  slits

            \beta is the fringe width which can also be evaluated as

                         \beta = \frac{L}{n}

Substituting values

                        \beta = \frac{0.0398}{24}

                          \beta = 1.660 *10^{-3}

Making d the subject of formula in the above equation

                d = \frac{\lambda D}{\beta }

Substituting values

                d = \frac{406.192 *10^{-9} * 5.937 }{1.660 *10^{-3}}

                    d = 0.00145m                

           

4 0
4 years ago
Determine la resistencia equivalente de la "escalera" de
defon

Las respuestas a cada inciso son:

a) La resistencia equivalente de la "escalera" de resistores iguales de 125 Ω que se muestra en la figura adjunta es:

R_{t} = 341.7 \: \Omega

b) La corriente a través de cada uno de los tres resistores de la izquierda si se conecta una batería de 50.0 V entre los puntos A y B es:

  • Correspondiente a R8 y R9 es 0.23 A
  • Correspondiente a R7 es 0.17 A.

a) En la imagen adjuntada correspondiente a la Figura 26-40, podemos observar que las resistencias 1, 2 y 3 están en serie, por lo tanto la ressitencia equivalente entre estas 3 es:

R' = R_{1} + R_{2} + R_{3}

De aquí en adelante tendremos presente que las todas las resistencias son iguales entre sí y por ende igual a 125 Ω. Las notaciones del 1 al 9 son para poder mostrar la resolución del problema.  

Entonces:

R' = 3R

Ahora, esta resistencia está en paralelo con la resistencia R₄, por lo tanto la resistencia equivalente entre estas dos es:

\frac{1}{R''} = \frac{1}{R'} + \frac{1}{R_{4}} = \frac{1}{3R} + \frac{1}{R} = \frac{4R}{3R^{2}}

R'' = \frac{3}{4}R

Luego, esta resistencia está en serie con las resistencias R₅ y R₆, por lo tanto:

R''' = R'' + R_{5} + R_{6} = \frac{3}{4}R + 2R = \frac{11}{4}R

Esta resistencia está ahora en paralelo con R₇, entonces:

\frac{1}{R''''} = \frac{1}{R'''} + \frac{1}{R_{7}} = \frac{4}{11R} + \frac{1}{R} = \frac{15R}{11R^{2}}

R'''' = \frac{11}{15}R

Finalmente, esta resistencis está en serie con las resistencias R₈ y R₉, por lo tanto la resistencia total es:

R_{t} = R'''' + R_{8} + R_{9} = \frac{11}{15}R + 2R = \frac{41}{15}R = \frac{41}{15}*125 \: \Omega  = 341.7 \: \Omega

b) Para este inciso debemos usar la Ley de Kirchhoff, pues tenemos tres mallas. Supondremos que las corriente de cada malla fluiran en sentido horario, por lo tanto las ecuaciones de para cada malla serán:

Malla 1

Segun la ley de Ohm tenemos:

V-i_{1}R-i_{3}R=0 (1)

Malla 2

-i_{2}R-i_{5}R-i_{2}R+i_{3}R=0 (2)

Malla 3

-i_{4}R-i_{4}R-i_{4}R+i_{5}R=0 (3)

Recordemos tambien que:

i_{1}=i_{2}+i_{3} (4)

i_{2}=i_{4}+i_{5} (5)

Lo que debemos hacer ahora es resolver el sistema de ecuaciones y encontrar los valore de las corrientes. Por lo tanto, los valores de las corrientes serán:

i_{1}=3/13\: A

i_{2}=4/65\: A

i_{3}=11/65\: A

i_{4}=1/65\: A

i_{5}=3/65\: A

     

Finalmente:

  • La corriente correspondiente a R8 y R9 es 0.23 A
  • La corriente correspondiente a R7 es 0.17 A.

Pudes aprender mas de mallas aquí:

https://brainly.lat/tarea/11593276

4 0
3 years ago
Rank the tensions in the ropes, t1, t2, and t3, from smallest to largest, when the boxes are in motion and there is no friction
gizmo_the_mogwai [7]
<span>AS T1,T2,T3 are the tensions in the ropes,assuming that there are Three blocks of mass 3m, 2m, and m.T3 is the string between 3m and 2m,T2 is the string between 2m and m ,T1 is the string attached to m thus T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.</span>
5 0
3 years ago
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