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lord [1]
3 years ago
9

The gravitational attraction between a 20 kg cannonball and a 0.002 kg

Physics
1 answer:
Naya [18.7K]3 years ago
3 0

Answer:

2.966\times 10^{-11}\ N

Explanation:

Given:

Mass of the cannonball (M) = 20 kg

Mass of the marble (m) = 0.002 kg

Distance between the cannonball and marble (d) = 0.30 m

Universal gravitational constant (G) = 6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2}

Now, we know that, the gravitational force (F) acting between two bodies of masses (m) and (M) separated by a distance (d) is given as:

F=\dfrac{GMm}{d^2}

Plug in the given values and solve for 'F'. This gives,

F=\frac{(6.674\times 10^{-11}\ m^3 kg^{-1} s^{-2})\times (20\ kg)\times (0.002\ kg)}{(0.30\ m)^2}\\\\F=\frac{6.674\times 20\times 0.002\times 10^{-11}\ m^3 kg^{-1+2} s^{-2}}{0.09\ m^2}\\\\F=2.966\times 10^{-11}\ kg\cdot m\cdot s^{-2}\\\\F=2.966\times 10^{-11}\ N.........(1\ N = 1\ kg\cdot m\cdot s^{-2})

The same force is experienced by both cannonball and marble.

Therefore, the gravitational  force of the marble is 2.966\times 10^{-11}\ N

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At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
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Answer:

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vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

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Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

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Big bang happened about 13.7 billion years ago in our universe.

<h3>Describe the beginning of the universe according to the big bang theory?</h3>

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An object accelerates from rest at a constant rate covering a distance of 18 meters in 3 seconds. What is its rate of accelerati
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3 years ago
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A volcanic eruption throws a boulder that lands 1.00 km horizontally from the crater. If the volcanic rocks were launched at an
Juliette [100K]

Answer:

a)  v₀ = 69.29 m / s , b)  t = 18.84 s

Explanation:

a) For this exercise we will use the projectile launch equations

          x = v₀ₓ t

          y = y₀ + v_{oy} t - ½ g t²

Let's fix our reference system on the volcano, so the horizontal distance x = 1 km = 1000 m and the vertical distance y = -900 m, the initial height of the crater is I = 0 m. Let's replace to find the speeds

         v_{oy} = v₀ sin θ

         v₀ₓ = v₀ cos θ

         y = v₀ sin θ (x / v₀ cos θ) - ½ g (x / v₀ cos θ)2

         y = x tan θ - ½ g x² / v₀² sec² θ

         ½ g x² sec² θ / v₀² = x tan θ - y

         v₀² = ½ g x² sec² θ / (x tan θ –y)

 

Let's calculate

           v₀² = ½ 9.8 1000² sec² 40 / (1000 tan 40 - (-900))

           v₀ = √ (8.35 10⁶ / 1,739 10³)

           v₀ = 69.29 m / s

    b) Flight time

           x = v₀ₓ t

           t = x / v₀ cos θ

           t = 1000 / 69.29 cos 40

           t = 18.84 s

7 0
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