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3 years ago

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A ball with an initial velocity of 8.00 m/s rolls up a hill without slipping. (a) Treating the ball as a spherical shell, calcul

ate the vertical height it reaches. (b) Repeat the calculation for the same ball if it slides up the hill without rolling.

1 answer:

GrogVix [38]3 years ago

3 0

Answer:

Part i)

h = 5.44 m

Part ii)

h = 3.16 m

Explanation:

Part i)

Since the ball is rolling so its total kinetic energy in this case will convert into gravitational potential energy

So we have

$\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = mgh$

here we know that for spherical shell and pure rolling conditions

$v = R \omega$

$I = \frac{2}{3}mR^2$

$\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{3}mR^2)(\frac{v^2}{R^2}) = mgh$

$\frac{5}{6}mv^2 = mgh$

$h = \frac{5v^2}{6g}$

$h = \frac{5(8^2)}{6(9.81)} = 5.44 m$

Part b)

If ball is not rolling and just sliding over the hill then in that case

$\frac{1}{2}mv^2 = mgh$

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

$h = \frac{8^2}{2(9.81)} = 3.16 m$

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Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.3 cm . Two of

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Answer:

$1.44\times 10^{-3}$ N

Explanation:

We are given that three charged particle are placed at each corner of equilateral triangle.

$q_1=-8.2 nC,q_2=-16.4 nC,q_3=8.0nC$

$q_1=-8.2\times 10^{-9} C$

$q_2=-16.4\times 10^{-9} C$

$q_3=8.0\times 10^{-9} C$

Side of equilateral triangle =3.3 cm= $\frac{3.3}{100}=0.033m$

We know that each angle of equilateral angle= $60^{\circ}$

Net force=F = $\sum\frac{kQq }{d^2}$

Where k= $9\times 10^9 Nm^2/C^2$

If we bisect the angle at $q_3$ then we have 30 degrees from there to either charge.

Direction of vertical force due to charge $q_1$ and $q_2$

Therefore, force will be added

Vertical force= $9\times 10^9\times q_3(q_1+q_2)\frac{cos30}{(0.033)^2})$

Vertical net force= $9\times 10^9\times 8\times 10^{-9}(-8.2-16.4)\times 10^{-9}\times 10^6\times\frac{\sqrt3}{2\times 1089}$

Vertical force = $9\times 8(-24.6)\times 10^{-9}\times 10^6\times 1.732\times \frac{1}{2178}$

Vertical force= $-1.41\times 10^{-3}N$ (towards $q_1}$

Horizontal component are opposite in direction then will b subtracted

Horizontal force= $9\times 10^9\times 8\times 10^{-9}(-8.2+16.4) \times 10^{-9}\times \frac{sin30}{(0.033)^2}$

Horizontal force= $0.27\times 10^-3}$ N(towards $q_2$

Net electric force acting on particle 3 due to particle = $\sqrt{F^2_x+F^2_y}$

Net force= $\sqrt{(-1.41\times 10^{-3})^2+(0.27\times 10^{-3})^2}$

Net force= $1.44\times 10^{-3}$ N

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Answer:

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12. A volcano erupts on Earth's surface causing a piece of rock with a mass of 95 kg to be ejected from rest to an acceleration

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Answer:

The force exerted on the rock by the eruption is, D. 902.5 N

Explanation:

Given data,

The mass of the rock ejected by the volcano, m = 95 kg

The acceleration of the ejected rock, a 9.5 m/s²

The force acting on an object is defined as the product of the mass and its acceleration. It is given by the relation,

F = m x a

= 95 x 9.5

= 902.5 N

Hence, the force exerted on the rock by the eruption is, F = 902.5 N

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2 years ago

Height of cannon 5 m, initial speed of projectile 15m/s, angle of launch 0 degrees. What is the range and time in the air? Pleas

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Answer:

<em>The range is 15.15 m and the time in the air is 1.01 s</em>

Explanation:

<u>Horizontal Motion</u>

When an object is thrown horizontally (with angle 0°) with a speed v from a height h, it follows a curved path ruled exclusively by gravity until it eventually hits the ground.

The range or maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

To calculate the time the object takes to hit the ground, we use the equation below:

$\displaystyle t=\sqrt{\frac{2h}{g}}$

The cannon is shot from a height of h=5 m with an initial speed of v=15 m/s. The range is calculated below:

$\displaystyle d=15\cdot\sqrt{\frac {2*5}{9.8}}=15*1.01$

d = 15.15 m

The time in the air is:

$\displaystyle t=\sqrt{\frac{2*5}{9.8}}$

t = 1.01 s

The range is 15.15 m and the time in the air is 1.01 s

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2 years ago

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

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Answer:

a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

Explanation:

a) From Second Newton's Law, we form this equation of equilibrium:

$\Sigma F = F_{E}-W = 0$ (Eq. 1)

Where:

$F_{E}$ - Electrostatic force exerted on human, measured in Newton.

$W$ - Weight of the human, measured in Newton.

If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:

$q\cdot E-m\cdot g = 0$

$q = \frac{m\cdot g}{E}$ (Eq. 2)

$E$ - Electric field, measured in Newtons per Coloumb.

$m$ - Mass, measured in kilograms.

$g$ - Gravity acceleration, measured in meters per square second.

$q$ - Electric charge, measured in Coulomb.

As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that $m = 60\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $E = -150\,\frac{N}{C}$ , the charge that a 60-kg human must have to overcome weight is:

$q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }$

$q = -3.923\,C$

The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.

b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:

$F = \kappa \cdot \frac{q^{2}}{r^{2}}$ (Eq. 3)

Where:

$\kappa$ - Electrostatic constant, measured in Newton-square meter per square Coulomb.

$q$ - Electric charge, measured in Coulomb.

$r$ - Distance between two people, measured in meters.

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q = -3.923\,C$ and $r = 100\,m$ , then the force of repulsion between two people is:

$F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right]$

$F = 13.851\times 10^{6}\,N$

The force of repulsion between two people is $13.851\times 10^{6}$ newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.

5 0

3 years ago