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Maksim231197 [3]
2 years 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]2 years 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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ioda

Answer:

6.1875\times 10^{-8}

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies I=\frac {k}{r^{2}} where r is the distance

Since intensity is given then when at 3 m

1.1\times 10^{-7}= \frac {k}{3^{2}}

k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}

Since we have the constant then at 4m

Intensity, I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}

8 0
2 years ago
A stone with mass 0.80 kg is attached to one end of a string 0.90 m long. The string will break if its tension exceeds 60.0 N. T
AleksandrR [38]

Answer:

v=8.2158m/s

Explanation:

(a) Free-body diagram attached.

(b) The stone attached with the string experiences both centripetal (towards the center) and centrifugal (away from the center) forces. The tension of the string counters the centrifugal force until it breaks.

We know that,

Centrifugal force = \frac{mv^2}{r}

where,

m = mass of the stone

v = velocity of the stone

r = length of the string

To find the maximum speed attained by the stone without the string breaking, we must equate:

\frac{mv^2}{r} =60

or, v=\sqrt \frac{{r\times 60}}{m} } =\sqrt{\frac{0.90\times60}{0.80} } =8.2158m/s

5 0
2 years ago
A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t
Morgarella [4.7K]

Answer:

(A). The work done is 1.50\times10^{-6}\ J.

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy K.E=1.50\times10^{-6}\ J

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

W = \Delta K.E

W=K.E_{f}-K.E_{i}

Put the value into the formula

W= 1.50\times10^{-6}-0

W=1.50\times10^{-6}\ J

The work done is 1.50\times10^{-6}\ J.

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

\Delta P.E=\Delta K.E

So, U = 1.50\times10^{-6}

Using formula of potential

V=\dfrac{U}{q}

Put the value into the formula

V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}

V=357.14\ V

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

W=F\times r....(I)

Using formula of force

F=qE

Put the value in the equation (I)

W=qE\times r

E=\dfrac{W}{q\times r}

Put the value into the formula

E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}

E=5952.38\ N/C

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

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Lena [83]
Disadvantage: <span>Living outside the city makes having a car almost a requirement. Businesses, schools, services, and recreational areas are all spread out at a distance, not to mention the need to commute to work.

Advantage: </span><span>The cost of a house with a yard in the suburbs is often less than the cost of a condominium or quality apartment in the city. </span>
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Yurem is pulling a wagon across the playground with a force of 10 N. He asks Elianna to help. She agrees and pushes the back of
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Answer:

22 N applied force

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