Preventing projects

A book is sliding to a stop while moving across the classroom floor

Margarita [4]

Answer:

can you please elaborate the question

Explanation:

3 0

2 years ago

Suppose you are standing on top of a hemisphere of radius r and you kick a soccer ball horizontally such that it has velocity v.

Ksivusya [100]

$|v| =\sqrt{ G \cdot M / r}$ , where

$M$ the mass of the planet, and
$G$ the universal gravitation constant.

Explanation:

Minimizing the initial velocity of the soccer ball would minimize the amount of mechanical energy it has. It shall maintain a minimal gravitational potential possible at all time. It should therefore stay to the ground as close as possible. An elliptical trajectory would thus be unfavorable; the ball shall maintain a uniform circular motion as it orbits the planet.

Equation 1 (see below) relates net force the object experiences, $\Sigma F$ to its orbit velocity $v$ and its mass $m$ required for it to stay in orbit :

$\Sigma F = m \cdot v^{2} / r$ (equation 1)

The soccer ball shall experiences a combination of gravitational pull and air resistance (if any) as it orbits the planet. Assuming negligible air resistance, the net force $\Sigma F$ acting on the soccer ball shall equal to its weight, $W = m \cdot g$ where $g$ the gravitational acceleration constant. Thus

$\Sigma F = W = m \cdot g$ (equation 2)

Substitute equation 2 to the left hand side of equation 1 and solve for $v$ ; note how the mass of the soccer ball, $m$ , cancels out:

$m \cdot g = \Sigma F = m \cdot v^{2} / r \\ v^{2} = g \cdot r \\ |v| = \sqrt{g \cdot r} \; (|v| \ge 0)$ (equation 3)

Equation 4 gives the value of gravitational acceleration, $g$ , a point of negligible mass experiences at a distance $r$ from a planet of mass $M$ (assuming no other stellar object were present)

$g = G \cdot M/ r^{2}$ (equation 4)

where the universal gravitation constant $G = 6.67408 \times 10^{-11} \cdot \text{m}^{3} \cdot \text{kg}^{-1} \cdot \text{s}^{-2}$

Thus

$\begin{array}{lll}|v| &=& \sqrt{g \cdot r}\\ & =&\sqrt{ G \cdot M / r}\end{array}$

3 0

3 years ago

(i) Calculate the energy, in electron volts, of a photon whose frequency is (a) 620 THz}, (b) 3.10GHz , and(c) 46.0 MHz

DedPeter [7]

The energy in electron volts of the photons that has the following frequencies is as follows:

620 THz = 2.564eV
3.10GHz = 1.28 × 10-⁵eV
46.0 MHz = 1.902 × 10-⁷eV

<h3>How to calculate energy?</h3>

The energy of a photon can be calculated using the following formula:

E = hf

Where;

E = energy
h = Planck's constant (6.626 × 10-³⁴ J/s)
f = frequency

First, we convert the frequencies to hertz as follows;

620THz = 6.2 × 10¹⁴Hz
3.10GHz = 3.1 × 10⁹Hz
46.0MHz = 4.6 × 10⁷Hz

E = 6.626 × 10-³⁴ × 6.2 × 10¹⁴ = 2.564eV
E = 6.626 × 10-³⁴ × 3.1 × 10⁹ = 1.28 × 10-⁵eV
E = 6.626 × 10-³⁴ × 4.6 × 10⁷ = 1.902 × 10-⁷eV

Learn more about energy of a photon at: brainly.com/question/2393994

#SPJ1

6 0

1 year ago

Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro

LUCKY_DIMON [66]

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

$\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}$

where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has

$\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}$

Again, we solve for the time it takes the ball to reach the ground:

$1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s$

After this time, we expect a horizontal displacement of 12 meters, so that $v_0$ satisfies

$v_0(0.57\,\mathrm s)=12\,\mathrm m$

$\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}$

5 0

3 years ago

Two stars that are in the same constellation:

Burka [1]

Answer:

Option (3)

Explanation:

A constellation is usually defined as a collection of stars that appears in the sky forming a particular type of shape or pattern. They are located at a very large distance from one another. These constellations are often used in order to locate any astronomical objects in space.

These stars appearing in a constellation generally appears to be closer to one another in the night sky, but they are actually located at about many light years away from one another.

For example, in the Orion constellation, the stars appearing in the sky are about 25-30 light-years away from one another.

Thus, the correct answer is option (3).

7 0

3 years ago

Preventing projects​

Preventing projects