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Maksim231197 [3]
1 year ago
7

Energy conservation with conservative forces: Two identical balls are thrown directly upward, ball A at speed v and ball B at sp

eed 2v, and they feel no air resistance. Which statement about these balls is correct?A) At their highest point, the acceleration of each ball is instantaneously equal tozero because they stop for an instant.B) The balls will reach the same height because they have the same mass and thesame acceleration.C) At its highest point, ball B will have twice as much gravitational potential energyas ball A because it started out moving twice as fast.D) Ball B will go twice as high as ball A because it had twice the initial speed.E) Ball B will go four times as high as ball A because it had four times the initialkinetic energy.
Physics
1 answer:
MatroZZZ [7]1 year ago
6 0

Answer:

E) True.   Ball B will go four times as high as ball A because it had four times the initial kinetic energ

Explanation:

To answer the final statements, let's pose the solution of the exercise

Energy is conserved

Initial

          Em₀ = K

          Em₀ = ½ m v²

Final

         Emf = U = mg h

         Em₀ = emf

        ½ m v² = mgh

        h = v² / 2g

For ball A

         h_A = v² / 2g

For ball B

        h_B = (2v)² / 2g

        h_B = 4 (v² / 2g) = 4 h_A

Let's review the claims

A) False. The neck acceleration is zero, it has the value of the acceleration of gravity

B) False. Ball B goes higher

C) False  has 4 times the gravitational potential energy than ball A

D) False.  It goes 4 times higher

E) True.

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Given the distance r = 2/1000 m, the force between them F = 0.0104 N, the mass of the two object can be calculated using formula:

F = G(m1m2)/r^2 since the mass are equal F = G (m^2)/r^2

And where G = is the gravitational constant (6.67E-11 m3 s-2 kg-1)

The mass of the two objects are 24.96 kg

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1 year ago
A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th
son4ous [18]

Answer:

Part a)

E = 8436.7 N/C

Part b)

E = 8436.7 N/C

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = \frac{Q}{A}

Q = -3.8 nC

A = \pi(0.09)^2

A = 0.025 m^2

\sigma = \frac{-3.8\times 10^{-9}}{0.025}

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now the electric field is given as

E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}

E = 8436.7 N/C

Part b)

Now since the electric field is required at same distance on other side

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E = 8436.7 N/C

5 0
2 years ago
Suppose a person has a small intestine that has fewer villi than normal. Would the person most likely be overweight or underweig
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They would be likely to be underweight. This is because the role of villi is to increase absorption of soluble molecules, they do this by increasing surface area for absorption to occur across.
If the person has less villi than normal in their small intestine, then the surface area will not be as large meaning there is less area for absorption to occur across so less soluble molecules will be absorbed.
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1 year ago
Assume that you stay on the earth's surface. what is the ratio of the sun's gravitational force on you to the earth's gravitatio
Pachacha [2.7K]
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg. 

F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N

Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
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Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
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5 0
1 year ago
Read 2 more answers
A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c
netineya [11]

Answer:

E=\frac{\lambda}{2\pi r\epsilon_0}

Explanation:

We are given that

Linear charge density of wire=\lambda

Radius of hollow cylinder=R

Net linear charge density of cylinder=2\lambda

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By Gauss theorem

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q=\lambda L

E(2\pi rL)=\frac{L\lambda}{\epsilon_0}

Where surface area of cylinder=2\pi rL

E=\frac{\lambda}{2\pi r\epsilon_0}

8 0
1 year ago
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