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qwelly [4]
3 years ago
6

George is planning an experiment to see how fast different sized cans of soup can roll down a ramp in his kitchen. He will be ro

lling them down a ramp and them time how fast it take them to travel 1 meter after the ramp. As George's Lab partner, what would be an appropiate hypothesis for George's Experiment?
a
The heavier cans will be faster, because science
b
If the can is denser, then it will take less time to cross 1 m, because gravity made it go faster down the ramp
c
The denser the soup can gets, the faster it will go, because gravity.
d
If the can is heavier, it will roll faster down the ramp due to gravity.
Physics
1 answer:
ddd [48]3 years ago
6 0
If it were me, I would choose D.
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Which is necessary for visualizing latent stains of blood, semen, or urine at a crime scene
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The correct answer is B) Ultraviolet light source. Hope this helps.
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3 years ago
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9. A 30 cm ruler is found to have a center of mass of 15.6 cm. The percent error of the center of mass is _____, if the ruler is
grigory [225]

Answer:

3.85 percent

Explanation:

From the question,

Percentage error = (error/actual)×100................ Equation 1

Given: actual center of mass = 15 cm, error = 15.6-15 = 0.6 cm

Substitute these values into equation 1

Percentage error = (0.6/15.6)×100

Percentage error = 3.85 percent

Hence the percentage error of the uniform mass = 3.85 percent

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3 years ago
If the frequency of q of a trait is 0.4, what is the frequency of p?
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In genetic traits, p and q represent the relative probabilities of the two alleles manifesting. If these two are the only options (ex. a dominant one and a recessive one), then the probabilities of both must sum up to 1. In this case, since we are given that q = 0.4, then p + q = 1, p + 0.4 = 1, and p = 0.6.
4 0
4 years ago
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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

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3 years ago
Name three mass units.! THANKS
erma4kov [3.2K]
Grams, Milligram, Microgram.
6 0
4 years ago
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