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Marat540 [252]
3 years ago
14

A ball is dropped off the roof of a tall building. If the ball reaches the ground in 8 seconds, how tall is the building, in met

ers? Round to the nearest whole number (a=9.8 m/s2)
Physics
1 answer:
Ghella [55]3 years ago
5 0

Let <em>b</em> be the height of the building, and <em>y</em> the height of the ball at time <em>t</em>, given by

<em>y</em> = <em>b</em> - 1/2 <em>gt</em>²

where <em>g</em> = 9.8 m/s² is the magnitude of the acceleration due to gravity.

It takes the ball 8 s to reach the ground, at which point <em>y</em> = 0, so that

0 = <em>b</em> - 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 1/2 (9.8 m/s²) (8 s)²

<em>b</em> = 313.6 m

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You have two small spheres, each with a mass of 2.40 grams, separated by a distance of 10.0 cm. You remove the same number of el
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Answer:

q = 2.066* 10⁻¹³ C.

n = 1,291,250 electrons.

Explanation:

1)

  • If the gravitational attraction is equal to their electrical repulsion, we can write the following equation:

       F_{g} = F_{c} (1)

  • where Fg is the gravitational attraction, that can be written as follows        according Newton's Universal Law of Gravitation:

       F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)

  • Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming  we can treat both spheres as point charges), as follows:

       F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)

  • since m₁ = m₂ = 0.0024 kg, and  r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

       G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)

  • Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

       Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)

  • Since both charges are the same, the charge on each sphere is just the square root of (5):
  • Q = 2.066* 10⁻¹³ C.

2)

  • Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
  • Now, since each electron carries an elementary charge equal to -1.6*10⁻¹⁹ C, in order to get the number of electrons removed from each sphere, we need to divide the charge removed from each sphere (the outcome of part 1) with negative sign) by the elementary charge, as follows:
  • n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)
4 0
3 years ago
The diagram below shows a golf ball being struck by a club. The ball leaves the club with a speed of 40 meters per second at an
hjlf

Answer:

61.22m

Explanation:

Maximum height in projectile is expressed as;

H = u²sin²θ/2g

u is the speed

g is the acceleration due to gravity

θ is the angle of projection

Substitute the given values into the formula;

H = u²sin²θ/2g

H = 40²sin²60/2g

H = 1600(0.8660)²/2(9.8)

H = 1600(0.8660)²/2(9.8)

H = 1,199.9/19.6

H = 61.22m

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A single point charge q is located at the center of both an imaginary cube and an imaginary sphere. How does the electric flux t
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Answer:

Explanation:

Electric flux is defined as the flow of electric field intensity through a given surface.

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\phi=E.A.cos\ \theta

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Electric flux through the surface of a sphere will be uniform throughout the surface area due to a charge at the center of the sphere. The distance of the surface from the center is always at a constant distance of radius of the sphere.

Electric flux through the surface of a cube will be varying as the surface area is at a varying distance from the center of the cube. The distance of the surface from the center is not at a uniform distance from the center of the cube and so the projection of solid angle changes.

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Answer:

Moving them farther apart

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E = \frac{K\times Q\times q}{r^{2}}

It shows that the electric field is inversely proportional to the square of the distance between two charges.

So, as the distance between two charges increases, the electric filed between the two charges decreases.

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