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MariettaO [177]
2 years ago
9

A spring is compressed so that it has 7.2 J of elastic potential energy. A 0.3 kg ball is placed on top of the spring. When the

spring is released, how high will the ball go ?
Physics
1 answer:
VLD [36.1K]2 years ago
5 0

The height of the ball is 2.45 m.

To calculate the height of the ball, we use the formula below.

<h3>Formula</h3>
  • E = mgh................ Equation 1

Where:

  • E = Elastic energy of the spring
  • m = mass of the ball
  • h = height of the ball
  • g = acceleration due to gravity.

make h the subject of the equation

  • h = E/mg...............Equation 2

From the question,

Given:

  • E = 7.2 J
  • m = 0.3 kg
  • g = 9.8 m/s²

Substitute these  values into equation 2

  • h = 7.2/(0.3×9.8)
  • h = 7.2/2.94
  • h = 2.45 m

Hence, The height of the ball is 2.45 m.

Learn more about height here: brainly.com/question/1739912


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Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

F=ma

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

8 0
2 years ago
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The force between two charges, q, and 92, is F. If the distance between the
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Answer:

4F

Explanation:

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F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

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A Red Rider bb gun uses the energy in a compressed spring to provide the kinetic energy for propelling a small pellet of mass 0.
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Answer:

a.6.5025 J

b.6.5025 J

Explanation:

We are given that

Mass of pellet,m=0.27 g=0.27\times 10^{-3} kg

1 kg=1000 g

Spring constant,k=1800 N/m

x=8.5 cm=8.5\times 10^{-2} m

1m=100 cm

a.Potential energy stored in the compressed spring  is given by

P.E=\frac{1}{2}kx^2

P.E=\frac{1}{2}(1800)(8.5\times 10^{-2})^2

P.E=6.5025 J

b.By using law of conservation of energy

P.E of spring=K.E of the pellet

K.E of the pellet=6.5025 J

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Một tải có điện trở R = 19ohm đấu vào nguồn điện một chiều có E = 100V,
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2 years ago
A tank contains 350 liters of fluid in which 10 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pu
aleksandr82 [10.1K]

Answer:

A(t) = -340e^{-t/70} + 350

Explanation:

Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.

Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t

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Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min

The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min

But the concentration is total amount of salt over 350L constant volume

C = A / 350

Therefore our rate of change for salt A' is

A' = 5 - 5A/350 = 5 - A/70

This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is

y = ce^{bt} + \frac{a}{b}

So A = ce^{\frac{-t}{70}} + \frac{5}{1/70} = ce^{-t/70} + 350

with A(0) = 10

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3 years ago
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