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Ugo [173]
3 years ago
14

Which dog has the most kinetic energy?

Physics
2 answers:
Nadya [2.5K]3 years ago
5 0

A. A dog of mass 14kg running with the speed of 3m/s

Furkat [3]3 years ago
3 0

Nobody actually needs to know how much kinetic energy any of these dogs has.  

The purpose of this question is to help you discover whether you know <em>how </em>to calculate kinetic energy, and then to give you some practice at <em>doing it</em>.

You have to calculate the kinetic energy of each dog, and then review your results to see which dog has the most.

The whole secret behind working this problem is the formula for the kinetic energy of any object:

<em>Kinetic Energy = (1/2) · (mass) · (speed)²</em>

The little ² next to 'speed' means 'squared' or 'multiplied by itself'.

(speed)² means (speed · speed) .

So, in order to calculate how much kinetic energy an object has, here's what you do:

-- Take the object's speed.  Multiply the speed by itself.

-- Take that number and multiply it by the object's mass.

-- Take that number and throw away half of it.

So now here we go:

Dog A:  KE = (1/2) (14 kg) (2 m/s)² = 28 Joules

Dog B:  KE = (1/2) (14 kg) (3 m/s)² = 63 Joules

Dog C:  KE = (1/2) (11 kg) (5 m/s)² = 137.5 Joules

Dog D:  KE = (1/2) 12 kg) (4 m/s)² = 96 Joules

==> <em>Dog C</em> has more kinetic energy than any of the other 3 dogs.

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3 years ago
The graph shows two runners participating in a race.
Olin [163]

Answer:

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Explanation:

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6 0
3 years ago
Read 2 more answers
The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f
Stells [14]

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

6 0
3 years ago
Each ball has a negligible size and a mass of 11.5 kg and is attached to the end of a rod whose mass may be neglected. The rod i
Digiron [165]

Answer:

3.31m/s

Explanation:

Angular momentum for 3s is

L = L_i_n_i + L_3_s

L = 2(11.5kg) + \int\limits^ {3s}_ {0s} {(t^2 + 2)} \, dt

L = 23kg+(\frac{t^3}{3} +2t)^ {3s}_ {0s}\\\\L=38kgm/s

Moment if inertia is

I = 2ml^2

I = 2(11.5)(0.5)^2\\\\I=5.75kgm^2

Angular speed

ω = L/I

= 38 / 5.75\\\\=6.61

The speed of each ball is

V = ωL

= 6.61\times0.5\\\\= 3.31m/s

7 0
3 years ago
What kind of frequency do radio waves have?<br><br>A. High frequency <br><br>B. Low frequency ​
inn [45]
B low frequency it is the lowest frequency
7 0
3 years ago
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