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marysya [2.9K]
3 years ago
7

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.

what is the minimum v that ensures the ball will not hit the hemispherical surface. in this case, how far does the ball land from the base of the hemisphere?
Physics
1 answer:
Ksivusya [100]3 years ago
3 0

|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.

<em>Equation 1</em>  (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 <em>(equation 1)</em>

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 <em>(equation 2)</em>

Substitute equation 2 to the left hand side of <em>equation 1</em> 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) <em>(equation 3)</em>

<em>Equation 4 </em> 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} <em>(equation 4)</em>

where the universal gravitation <em>constant</em> 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}

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A 10 kg frictionless cart is pushed at a constant force of 5.0 N for a distance of 10 m. The work done on the cart is 50J.
kykrilka [37]

1) Final kinetic energy of the cart: B) 50 J

2) Final speed of the cart: C) 3.2 m/s

3) Height reached along the ramp: A) 0.5 m

Explanation:

1)

We can solve this part of the problem by using the work-energy theorem, which states that the work done on an object is equal to the kinetic energy gained by the object itself. Mathematically:

W=K_f - K_i

where

W is the work done

K_f is the final kinetic energy

K_i is the initial kinetic energy

In this problem, the work done on the cart is

W = 50 J

And assuming it starts from rest, its initial kinetic energy is zero:

K_i = 0

Therefore, the final kinetic energy is:

K_f = K_i + W=0+50=50 J

2)

The kinetic energy of an object is the energy possessed by an object due to its motion; it is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the cart in this problem, we have:

K = 50 J is its final kinetic energy

m = 10 kg is the cart

Therefore, solving the formula for v, we find its speed:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(50)}{10}}=3.2 m/s

3)

We can think this problem in terms of conservation of energy. In fact, as the cart rolls up the ramp, its kinetic energy is converted into gravitational potential energy, which is given by

PE=mgh

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

h is the heigth of the cart

When the cart reaches the maximum height, all the kinetic energy has been converted into potential energy, so we can write:

K=PE\\\frac{1}{2}mv^2=mgh

Re-arranging,

h=\frac{v^2}{2g}

And since we know the initial speed of the cart along the ramp,

v = 3.2 m/s

we can find the maximum height reached along the ramp:

h=\frac{3.2^2}{2(9.8)}=0.5 m

Learn more about work, kinetic energy and potential energy:

brainly.com/question/6763771

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brainly.com/question/6536722

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#LearnwithBrainly

7 0
3 years ago
6. What is the speed (v) of an object with 13 kg of mass and a momentum of 3 kg-<br> m/s?
I am Lyosha [343]

Answer:

0.23m/s

Explanation:

momentum(p) = mv

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7 0
3 years ago
A uniform, solid, 100-kg cylinder with a diameter of 1.0 m is mounted so it is free to rotate about fixed, horizontal, frictionl
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Answer:

Explanation:

The cylinder effective mass is one half of its actual mass

F = ma

10(9.8) = (10 + 100/2)a

a = 1.633333... ≈ 1.6 N

T = m(g - a)

T = 10(9.8 - 1.6) = 8.2 N

y = ½at²

3.2 = ½(1.6)t²

t = 2.0 s

If one wants to go from basics, The FBD on the cylinder is

τ = Iα

TR = ½MR²α

T = ½MRα

T = ½MR(a/R)

T = ½Ma

The FBD on the block with down being positive is

mg - T = ma

substituting from the cylinder analysis

mg - ½Ma = ma

mg = a(m + ½M)

a = mg / (m + ½M)    which is identical to the equation I used originally.

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3 years ago
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