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3 years ago

A 60kg student is 5m from a 120kg door, how much gravitational forces act between the two?

Physics

1 answer:

slavikrds [6]3 years ago

7 0

Gravitational force = G · m₁ · m₂ / r²

' G ' is the 'universal gravitational constant, 6.673 x 10⁻¹¹ m³/kg-s²

m₁ = 60 kg

m₂ = 120 kg

r = 5 m

F = (6.673 x 10⁻¹¹ m³/kg-sec²) · (60 kg) · (120 kg) / (5m)²

F = (6.673 x 10⁻¹¹ · 60 · 120 / 25) · (m³ · kg · kg / kg · s² · m²)

F = 1.922 x 10⁻⁸ (kg · m / s²)

<em>F = 1.922 x 10⁻⁸ Newton</em> (about 0.0000000692 ounce of force)

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Cobaltâ’60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this

Crank

Energy of gamma rays is given by equation

$E = h\nu$

here we know that

h = Planck's constant

$\nu = frequency$

now energy is given as

$E = 4.70 MeV = 4.70 \times 10^6 \times 1.6 \times 10^{-19}$

$E = 7.52 \times 10^{-13} J$

now by above equation

$E = h\nu$

$7.52 \times 10^{-13} = 6.6 \times 10^{-34} \nu$

$\nu = 1.14 \times 10^{21} Hz$

now for wavelength we can say

$\lambda = \frac{c}{\nu}$

$\lambda = \frac{3\times 10^8}{1.14 \times 10^{21}}$

$\lambda = 2.63 \times 10^{-13} m$

3 0

2 years ago

A proton is released in a uniform electric field, and it experiences an electric force of 2.04×10−14 N toward the south.

Lemur [1.5K]

Answer:

Part A:

$E=127500N/C\\E=1.275*10^{5} N/C$

Part B:

Option B (Towards the South)

Explanation:

Part A:

Magnitude if electric field E:

E=Force/charge

Force=2.04×10−14 N

Charge=1.6×10−19 C

$E=\frac{2.04*10^{-14}}{1.6*10^{-19}} \\E=127500N/C\\E=1.275*10^{5} N/C$

Part B:

Option B (Towards the South)

As electron is experiencing the force towards south,it means the direction of the electric field is towards the south because direction of field lines is from positive to negative, so proton is moving towards south it means negative charge is in south to which proton is attracted. So electric field is towards South.

7 0

3 years ago

Which road would exert the LEAST amount of friction on a car?

bearhunter [10]

Answer:

Icy roads

Explanation:

There is so little friction you slide on it way more than other roads. :)

6 0

2 years ago

un movil que parte del reposo alcanza una velocidad de 75 m/s en 13 segundos ¿cual su aceleracion y el espacio que recorrio en l

Dmitriy789 [7]

Answer:

Acceleration = 5.77 m/s²

Distance cover in 13 seconds = 487.56 meter

Explanation:

Given:

Final velocity of mobile device = 75 m/s

initial velocity of mobile device = 0 m/s

Time taken = 13 seconds

Find:

Acceleration

Distance cover in 13 seconds

Computation:

v = u + at

75 = 0 + (a)(13)

13a = 75

a = 5.77

Acceleration = 5.77 m/s²

s = ut + (1/2)(a)(t²)

s = (0)(t) + (1/2)(5.77)(13²)

Distance cover in 13 seconds = 487.56 meter

8 0

2 years ago

A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical

svp [43]

Answer:

X-Positions: Y-Positions

x(0) = 0 y(0) = 0

x(2) = 120 m y(2) = 19.6 m

x(4) = 240 m y(4) = 78.4 m

x(6) = 360 m y(6) = 176.4 m

x(8) = 480 m y(8) = 313 m

x(10) = 600m y (10) = 490 m

Explanation:

X-Positions

First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

$x = v_{ox} * t = 60.0 m/s * t(s)$

It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
At any moment, it is subject to the acceleration of gravity, g.
As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:

$y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}$

Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
y(2) = 2* 9.8 m/s2 = 19.6 m
y(4) = 8* 9.8 m/s2 = 78.4 m
y(6) = 18*9.8 m/s2 = 176.4 m
y(8) = 32*9.8 m/s2 = 313.6 m
y(10)= 50 * 9.8 m/s2 = 490.0 m

5 0

3 years ago