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ivolga24 [154]
3 years ago
7

Gravity causes objects to be attracted to one another. This attraction keeps our feet firmly planted on the ground and causes th

e moon to orbit the earth. The force of gravitational attraction is represented by the equation (drawn) where F is the magnitude of the gravitational attraction on either body,m1 and m2 are the masses of the bodies, r is the distance between them, and G is the gravitational constant. In SI units, the units or force are kg m/s2, the units of mass are kg, and the units of distance are m. For this equation to have consistent units, the units of G
Physics
1 answer:
Airida [17]3 years ago
5 0
F = G m1*m2 / r^2 => [G] = [F]*[r]^2 /([m1]*[m2]) = N * m^2 / kg^2

That is one answer.

Also, you can use the fact that N = kg*m/s^2

[G] = kg * m / s^2 * m^2 / kg^2 = m^3 /(s^2 * kg)
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Dos cargas iguales de 10^-7C estan separadas por una distancia de 2m. calcule la fuerza con que se repelen
m_a_m_a [10]

Answer:

F=2.25\times 10^{-5}\ N

Explanation:

According to question,

Charge 1 and charge 2 are 10^{-7}\ C

The distance between charges is 2 m

We need to find the force with which two positive charges repel. It is called electrostatic force of repulsion. It can be given by :

F=\dfrac{kq^2}{r^2}\\\\F=\dfrac{9\times 10^9\times (10^{-7})^2}{2^2}\\\\F=2.25\times 10^{-5}\ N

So, the electric force of repulsion is 2.25\times 10^{-5}\ N.

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3 years ago
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G a person of mass 100 kg is riding an elevator which was initially moving up with a velocity of 3 m/s. over a distance of 4 m t
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3 0
2 years ago
1. Bone has a Young’s modulus of about
Blababa [14]

#1

As we know that

Y = \frac{stress}{strain}

now plug in all data into this

1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}

strain = 9.33 \times 10^{-3}

now from the formula of strain

strain = \frac{\Delta L}{L}

9.33 \times 10^{-3} = \frac{\Delta L}{0.54}

\Delta L = 5.04 \times 10^{-3} m

\Delta L = 5.04 mm

#2

As we know that

pressure * area = Force

here we know that

Area = 3.53 \times 11.6 = 40.95 m^2

P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa

now force is given as

F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N

#3

As we know that density of water will vary with the height as given below

\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}

here we know that

\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa

B = 2.3 \times 10^9 N/m^2

now density is given as

\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}

\rho = 1185.3 kg/m^3

#4

as we know that pressure changes with depth as per following equation

P = P_o + \rho g h

here we know that

P = 3 P_0

now we will have

3P_0 = P_0 + \rho g h

2P_0 = \rho g h

2(1.01 \times 10^5) = 1025 (9.81)(h)

here we will have

h = 20.1 m

so it is 20.1 m below the surface

#5

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

mg = \rho_1V_1g + \rho_2V_2g

A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)

46.6 = 43.89 - 922x + 1000x

x = 3.48 cm

so it is 3.48 cm below the interface

5 0
3 years ago
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