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3 years ago

What is the potential energy of the bowling ball as it sits on top of a building ?

Physics

1 answer:

tangare [24]3 years ago

8 0

Answer:

The potential energy of the bowling ball will be mgh

Explanation:

Let the mass of bowling ball =m

The height of building on which bowling ball sits=h

So,

The potential energy of the bowling ball =P.E.= mgh

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Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of +6q. Sphere B caries a charge of-2q. Sphere C

miskamm [114]

<h2>20. How much charge is on sphere B after A and B touch and are separated?</h2><h3>Answer:</h3>

$\boxed{q_{B}=+2q}$

<h3>Explanation:</h3>

We'll solve this problem by using the concept of electric potential or simply called potential $V$ , which is <em>the energy per unit charge, </em>so the potential $V$ at any point in an electric field with a test charge $q_{0}$ at that point is:

$V=\frac{U}{q_{0}}$

The potential $V$ due to a single point charge q is:

$V=k\frac{q}{r}$

Where $k$ is an electric constant, $q$ is value of point charge and $r$ is the distance from point charge to where potential is measured. Since, the three spheres A, B and C are identical, they have the same radius $r$ . Before the sphere A and B touches we have:

$V_{A}=k\frac{q_{A}}{r_{A}} \\ \\ V_{B}=k\frac{q_{B}}{r_{A}} \\ \\ But: \\ \\ \ r_{A}=r_{B}=r$

When they touches each other the potential is the same, so:

$V_{A}= V_{B} \\ \\ k\frac{q_{A}}{r}=k\frac{q_{B}}{r} \\ \\ \boxed{q_{A}=q_{B}}$

From the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant. </em>So:

$q_{A}+q_{B}=q \\ \\ q_{A}=+6q \ and \ q_{B}=-2q \\ \\ So: \\ \\ \boxed{q_{A}+q_{B}=+4q}$

Therefore:

$(1) \ q_{A}=q_{B} \\ \\ (2) \ q_{A}+q_{B}=+4q \\ \\ (1) \ into \ (2): \\ \\ q_{A}+q_{A}=+4q \therefore 2q_{A}=+4q \therefore \boxed{q_{A}=q_{B}=+2q}$

So after A and B touch and are separated the charge on sphere B is:

$\boxed{q_{B}=+2q}$

<h2>21. How much charge ends up on sphere C?</h2><h3>Answer:</h3>

$\boxed{q_{C}=+1.5q}$

<h3>Explanation:</h3>

First: A and B touches and are separated, so the charges are:

$q_{A}=q_{B}=+2q$

Second: C is then touched to sphere A and separated from it.

Third: C is to sphere B and separated from it

So we need to calculate the charge that ends up on sphere C at the third step, so we also need to calculate step second. Therefore, from the second step:

Here $q_{A}=+2q$ and C carries no net charge or $q_{C}=0$ . Also, $r_{A}=r_{C}=r$

$V_{A}=k\frac{q_{A}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

Applying the same concept as the previous problem when sphere touches we have:

$k\frac{q_{A}}{r} =k\frac{q_{C}}{r} \\ \\ q_{A}=q_{C}$

For the principle of conservation of charge:

$q_{A}+q_{C}=+2q \\ \\ q_{A}=q_{C}=+q$

Finally, from the third step:

Here $q_{B}=+2q \ and \ q_{C}=+q$ . Also, $r_{B}=r_{C}=r$

$V_{B}=k\frac{q_{B}}{r} \\ \\ V_{C}=k\frac{q_{C}}{r}$

When sphere touches we have:

$k\frac{q_{B}}{r} =k\frac{q_{C}}{r} \\ \\ q_{B}=q_{C}$

For the principle of conservation of charge:

$q_{B}+q_{C}=+3q \\ \\ q_{A}=q_{C}=+1.5q$

So the charge that ends up on sphere C is:

$q_{C}=+1.5q$

<h2>22. What is the total charge on the three spheres before they are allowed to touch each other.</h2><h3>Answer:</h3>

$+4q$

<h3>Explanation:</h3>

Before they are allowed to touch each other we have that:

$q_{A}=+6q \\ \\ q_{B}=-2q \\ \\ q_{C}=0$

Therefore, for the principle of conservation of charge <em>the algebraic sum of all the electric charges in any closed system is constant, </em>then this can be expressed as:

$q_{A}+q_{B}+q_{C}=+6q -2q +0 \\ \\ \therefore q_{A}+q_{B}+q_{C}=+4q$

Lastly, the total charge on the three spheres before they are allowed to touch each other is:

$+4q$

8 0

3 years ago

A baseball rolls off a 1.20m high desk and strikes the floor 0.50m away from the base of the desk . How fast was it rolling?

noname [10]

The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration ( $g=9.8 m/s^2$ , acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

$s=ut+\frac{1}{2}gt^2$

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

$g=9.8 m/s^2$

t is the time of flight

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s$

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

in a time

t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

$v_x = \frac{d}{t}=\frac{0.50}{0.495}=1.01 m/s$

So, the ball rolls off the table at 1.01 m/s.

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

4 0

3 years ago

. A solenoid coil consists of a single layer of 250 circular turns of wire with each turn having a 0.02m radius. The axial lengt

Vinil7 [7]

Answer:

a) L = 3.29 10⁻⁴ H, b)U = 5.33 10⁻² J

Explanation:

a) The inductance is a solenoid this given carrier

L = $\frac{N \ \phi_B }{I}$

The magnetic field inside the solenoid is

B = μ₀ $\frac{N}{l} i$

hence the magnetic flux

Ф_B = B. A = μ₀ $\frac{N \ A}{l \ i}$

we substitute in the expression of inductance

L = N² μ₀ A /l

let's find the area of each turn

A = π r²

A = π 0.02²

A = 1.2566 10⁻³ m²

let's calculate

L = 250² 4π 10⁻⁷ 1.2566 10⁻² / 0.3

L = 3.29 10⁻⁴ H

b) The stored energy is

U = ½ L i²

let's calculate

U = ½ 3.29 10⁻⁴ 18²

U = 5.33 10⁻² J

5 0

3 years ago

Q.By the method of dimensional analysis derive the

Artyom0805 [142]

Answer

a) Using dimensional analysis we cannot derive the relation, But we can check the correctness of the formula.

$s = u t +\dfrac{1}{2}at^2$

now, L H S

s = distance

dimension of distance = [M⁰L¹T⁰]

now, equation on the right hand side

R H S

u = speed

u = m/s

Dimension of speed = [M⁰L¹T⁻¹]

dimension of time

t = sec

Dimension of time = [M⁰L⁰T¹]

Dimension of 'ut' = [M⁰L¹T⁻¹][M⁰L⁰T¹]

= [M⁰L¹T⁰]

now, acceleration= a

a = m /s²

dimension of acceleration = [M⁰L¹T⁻²]

dimension of (at²) = [M⁰L¹T⁻²][M⁰L⁰T¹][M⁰L⁰T¹]

= [M⁰L¹T⁰]

hence, the dimension are balanced.

so, L H S = R H S

b) Moment of inertia of hollow sphere = $\dfrac{2}{3}Mr^2$

Moment of inertia of solid sphere = $\dfrac{2}{5}Mr^2$

we know,

$\tau = I \alpha$

$\alpha=\dfrac{\tau}{I}$

Torque is the force that causes rotation

If the same amount of torque is applied to both spheres the sphere with bigger moment of inertia would have smaller angular velocity.

Thus the solid sphere would accelerate more.

4 0

3 years ago

Read 2 more answers

2. An object of mass 20 kg is lifted to a 25 m building. How much potential energy is stored on the mass? (Take g= 10 m/s)

butalik [34]

Explanation:

potential energy = mass × gravity force × height

=20 × 10 × 25

= 5000 joules

6 0

2 years ago

What is the potential energy of the bowling ball as it sits on top of a building ?​

What is the potential energy of the bowling ball as it sits on top of a building ?