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Paha777 [63]
2 years ago
5

Goku vs saitama idrc what the answer is

Physics
2 answers:
Naya [18.7K]2 years ago
7 0

Answer:

goku vs saitama idrc what the answer is

Saitama wins

weqwewe [10]2 years ago
4 0

Answer:

it isnt a fair fight (tie)

Explanation:

in Saitama's world, he does Push-ups – 100. Sit-ups – 100. Squats – 100. Running (10 km around 6.1 miles) The same process goes on for 7 days a week. but in one episode of dragon ball super goku did one trillion situps without trying so this means if goku went to saitamas world he would be a one punch man but id saitamas went to gokus world saitama would become a god and train like goku this means atmosphere is differnt  so if  they crossed paths whoever is in their home planet would win bc of the atmosphere                    

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An object is placed 50.0 cm in front of a convex mirror. where can be the image located if the focal length is 40 cm from the mi
sergiy2304 [10]

Answer:

Image will form at distance 22.22 cm behind the mirror

Explanation:

As we know that the mirror formula is given as

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

now we know that

object distance from mirror is

d_o = -50 cm

Focal length of the mirror is given as

f = 40 cm

now we have

\frac{1}{-50} + \frac{1}{d_i} = \frac{1}{40}

\frac{1}{d_i} = \frac{1}{50} + \frac{1}{40}

d_i = 22.22 cm

6 0
3 years ago
EXTREME HELP!!!!
mrs_skeptik [129]

the anwser is c is always the best anwser


8 0
3 years ago
Si el coeficiente de fricción cinética entre los neumáticos y el pavimento seco es de 0.80. ¿Cuál es la distancia mínima para de
vovangra [49]

Answer: 52.9 metros.

Explanation:

Podemos escribir la fuerza de fricción cinética como

F = μ*N

donde N es la fuerza normal entre el coche y el suelo, cuya magnitud es igual al peso en esta situación.

F = μ*m*g

donde m es la masa del coche y g es 9.8m/s^2

y sabemos que μ = 0.8

Por la segunda ley de Newton, sabemos que:

F = m*a

fuerza es igual a masa por aceleración.

a = F/m

entonces la aceleración causada por la fuerza de rozamiento es:

F = 0.8*m*g

a = F/m = (0.8*m*g)/m = 0.8*g.

Entonces ya encontramos la aceleración, hay que recordar que esta aceleración es en sentido opuesto a la sentido de movimiento, entonces podemos escribir la aceleración como:

a(t) = -0.8*g

Para la velocidad, podemos integrar sobre el tiempo para obtener.

v(t) = -0.8*g*t + v0

donde v0 es la velocidad inicial del auto = 28.7m/s

v(t) = -0.8*g*t + 28.8m/s

Ahora podemos encontrar el tiempo necesario para que la velocidad del coche sea cero, en ese momento, como deja de moverse, ya no tendremos rozamiento cinético, entonces no habrá aceleración y el coche se detendrá completamente.

v(t) = 0m/s = -0.8*9.8m/s^2*t + 28.8m/s

7.84m/s^2*t = 28.8m/s

                 t   = (28.8m/s)/(7.84m/s^2) = 3.63 segundos.

Ahora vamos a la ecuación de movimiento, donde asumimos que la posición inicial del coche es 0m, así que no tendremos constante de integración.

p(t) = -(1/2)*(0.8*9.8m/s^2)*t^2 + 28.8m/s*t

Ahora podemos evaluar la posición en t = 3.63 segundos, y esto nos dara la distancia que el coche se movio mientras frenaba.

p(3.63s) = -(1/2)*(0.8*9.8m/s^2)*(3.63s)^2 + 28.8m/s*(3.63s) = 52.9 metros.

6 0
3 years ago
A glass bottle of soda is sealed with a screw cap. The absolute pressure of the carbon dioxide inside the bottle is 1.50 x 105 P
MArishka [77]

Answer:

F \approx 19.5 N

Explanation:

From the question we are told that:

Pressure of  P_{CO_2}=1.50 * 105 Pa.

Bottle cap area A_b= 4.40 * 10-4 m^2

 

Generally the equation for Resultant pressure P_r is give as is mathematically given by

P_r=P_{CO_2}-P_a

Where

P_a=atmospheric\ pressure = 1.013*10^5 pa

P_r=1.50 * 105 Pa-1.013*10^5 pa

P_r=0.487*10^5 pa

Generally the equation for Force exerted by screw F is give as is mathematically given by

F = P*A\\F = 0.487*10^5*4.00*10^-4\\ F = 19.48 N

F \approx 19.5 N

4 0
2 years ago
Student 1 lifts a box with a force of 500 N and sets it on a tabletop 1.2 m high. Student 2 pushes an identical box up a 5 m ram
Troyanec [42]

The student who did the most work is student 2 with 2500 Joules.

<u>Given the following data:</u>

  • Force 1 = 500 Newton
  • Distance 1 = 1.2 meter
  • Force 2 = 500 Newton
  • Distance 2 = 5 meter

To determine which of the students did the most work:

Mathematically, the work done by an object is given by the formula;

Work\;done = Force \times distance

<u>For </u><u>student 1</u><u>:</u>

Work\;done = 500 \times 1.2

Work done = 600 Joules

<u>For </u><u>student 2</u><u>:</u>

Work\;done = 500 \times 5

Work done = 2500 Joules.

Therefore, the student who did the most work is student 2 with 2500 Joules.

Read more: Read more: brainly.com/question/13818347

7 0
3 years ago
Read 2 more answers
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