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2 years ago

How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m? (The acceleration due to gravity is 9.8 m/s2.)

Physics

2 answers:

V125BC [204]2 years ago

7 0

<span>How much kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m?
</span>Answer:

85 Joule (approx)

Explanation:

Potential energy at highest point

<span><span>P.E = mgh = 7.2 kg × 9.8 m/s2</span>×1.2 m≈85 Joule</span>

This is the kinetic energy required for dog to jump a height of 1.2 m.

hope this helps!

Tasya [4]2 years ago

6 0

Answer:

Kinetic energy of dog is 84.67 Joules

Explanation:

Given that,

Mass of the dog, m = 7.2 kg

We have to find the kinetic energy does a 7.2-kg dog need to make a vertical jump of 1.2 m. We know that the energy of a system remains conversed. At highest point there is only potential energy. So, while taking a vertical jump the potential energy gets converted to potential energy.

So, $E_k=mgh$

here, h = 1.2 m

$E_k=7.2\times 9.8\times 1.2$

$E_k=84.67\ Joules$

Hence, the kinetic energy needed to make a vertical jump is 84.67 Joules.

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Nikolay [14]

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3 years ago

A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3 Pa s and

icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

g = acceleration due to gravity = 9.8 m/s²,
r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
ρ = density of sphere = 19 g/cm³,
σ = density of water = 1.0 g/cm³ and
η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

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1 year ago

What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures

Montano1993 [528]

To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.

Ff = 0.5 * 16 * 9.8 = 78.4 N

a = 4.9 m/s^2

If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?

In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.

Ff = 0.28 * 16 * 9.8 = 43.904 N

Net force = 78.4 – 43.904 = 34.496 N

To determine the acceleration, divide by the mass of the crate.

a = 34.496 ÷ 16 = 2.156 m/s^2

8 0

3 years ago

A worker pushes a box across the floor to the right at a constant speed with a force of 25N. What

icang [17]

Answer:

Friction between the box and the floor is 25N to the left

Explanation:

There are two forces acting on the box along the horizontal direction:

- The force of push applied by the worker, in the forward direction, F

- The force of friction, $F_f$ , acting in the opposite direction (backward)

So the net force acting on the box is

$F_{net}=F-F_f$

According to Newton's second law of motion, the net force on an object is equal to the product between its mass (m) and its acceleration (a), so we can write:

$F_{net}=ma$