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3 years ago

A marble rolls with a speed of 15 m/s and has a momentum of 0.15 kg*m/s. What is its mass?

Physics

1 answer:

prisoha [69]3 years ago

5 0

Answer:

m = 0.01 kg

Explanation:

Given that,

Momentum of the marble, p = 0.15 kg-m/s

Speed of the marble, v = 15 m/s

We need to find its mass. We know that,

Momentum, p = mv

Where

m is the mass

$m=\dfrac{p}{v}\\\\m=\dfrac{0.15}{15}\\\\m=0.01\ kg$

So, the mass of the marble is equal to 0.01 kg.

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Rutherford's gold-foil experiment led him to conclude that

erica [24]

Answer:

C. a dense region of positive charge existed somewhere in the atom.

Explanation:

Physicist Ernest Rutherford created the gold foil experiment in which he shot a beam of alpha particles at a sheet of gold foil, which then sent a few of the particles flying after they were deflected. Based on the information gathered after completing this experiment, Rutherford concluded that a dense region of positive charge existed somewhere in the atom.

I hope this answered your question. If you have any more questions feel free to ask away at Brainly.

3 0

3 years ago

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

An expensive vacuum system can achieve a pressure as low as 1.00×10^{-7} N/m^2 at 20ºC . How many atoms are there in a cubic cen

marta [7]

Answer:

$24.70818432141\times 10^7\ atoms$

Explanation:

P = Pressure = $1\times 10^{-7}\ N/m^2$

V = Volume = 1 cm³

n = Amount of substance

N = Number of atoms

$N_A$ = Avogadro's constant = $6.022\times 10^{23}\ /mol$

R = Gas constant = 8.314 J/k mol

T = Temperature = 273.15+20 = 293.15 K

From the ideal gas law

$PV=nRT\\\Rightarrow n=\frac{PV}{RT}$

$n=\frac{N}{N_A}$

$\frac{N}{N_A}=\frac{PV}{RT}\\\Rightarrow N=N_A\times \frac{PV}{RT}\\\Rightarrow N=\frac{1\times 10^{-7}\times 1\times 10^{-6}}{8.314\times 293.15}\times 6.022\times 10^{23}\\\Rightarrow N=24708184.32141\ atoms=24.70818432141\times 10^7\ atoms$