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3 years ago

Not a assignment question, just wanna know if scientists know what’s beyond the edge universe

2 answers:

8 0

No they don’t. All the pictures we see if the universe are computer generated. Well, most of them are. We don’t exactly know how the milky way looks. It’s just very good guess.

Oksanka [162]3 years ago

6 0

I don’t think so because we can even get to mars yet how are we supposed to know what’s beyond our universe like i understand beyond solar system but not our universe

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Imagina que compras una placa rectangular de metal de 2mm de alto, 10mm de ancho x 50mm de largo, y una masa de 0.02kg. El vende

shusha [124]

Answer:

Densidad de la placa = 20 g/cm³.

La placa no es de oro.

Explanation:

Para encontrar la densidad de la placa rectangular primero debemos hallar su volumen:

$V = 2 mm*10 mm*50 mm = 1000 mm^{3}*\frac{1 cm^{3}}{(10 mm)^{3}} = 1 cm^{3}$

Ahora, encontremos al densidad de la placa:

$d = \frac{m}{V} = \frac{20 g}{1 cm^{3}} = 20 g/cm^{3}$

Dado que la densidad del oro es 19.32 g/cm³ y que la densidad de la placa rectangular calculada es 20 g/cm³, podemos decir que dicha placa no es de oro.

Espero que te sea de utilidad!

6 0

3 years ago

A race car accelerates from 0 m/s to 30.0 m/s with a displacement of

worty [1.4K]

Answer:

4. 10.0 m/s²

Explanation:

I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:

$1. \ a=\frac{v-v_0}{t};$

$2. \ L=\frac{at^2}{2}.$

II) using these two equations after substitution v₀=0; v=30 and L=45:

$\left \{{{45 =\frac{at^2}{2}} \atop {a=\frac{30-0}{t} }} \right.$

$\left \{ {{at^2=90} \atop {at=30}} \right. \ \ \left \{ {{a=10} \atop {t=3}} \right.$ $=> \ a=10\frac{m}{s^2}$

6 0

3 years ago

What type of stress is caused by two tectonic plates sliding past one another

prisoha [69]

Extensional stress. is your answer.

5 0

3 years ago

Read 2 more answers

Why do scientist use different types of models to represent compounds

nexus9112 [7]

Because they are different they all show different traits.

5 0

3 years ago

A wire lying along a y axis form y=0 to y=0.25 m carries a current of 2.0 mA in the negative direction of the axis. The wire ful

balu736 [363]

Answer:

FB = 0.187 N

Explanation:

To find the magnetic force FB in the wire you use the following formula:

$|\vec{F_B}|=ILBsin\theta\\\\L=0.25m\\\\|\vec{B}|=\sqrt{(0.3y)^2+(0.4y)^2}=0.5y \ T$

the angle between B and L is given by:

$\vec{L}\cdot\vec{B}=LBcos\theta\\\\\theta=cos^{-1}(\frac{\vec{L}\cdot\vec{B}}{LB})=cos^{-1}(\frac{0*0.3y+0.25*0.4y}{0.25*0.5y})=36.86\°$

Due to B depends on "y" you take into account the contribution of each element dy of the wire to the magnitude of the magnetic force. Thus, you have to integrate the following expression:

$|\vec{F_B}|=Isin\theta\int_0^{0.25}B(y)dy=Isin\theta\int_0^{0.25}(0.5y)dy\\\\|\vec{F_B}|=(2.0*10^{-3}A)(sin36.86\°)(0.5T)[\frac{0.25^2}{2}m]=0.187\ N$

hence, the magnitude of the magnetic force is 0.187N

4 0

3 years ago