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2 years ago

Determine whether each object has potential energy, kinetic energy, or both

Physics

1 answer:

Pepsi [2]2 years ago

4 0

Answer:

stretched spring has potential energy
a falling skydiver has kinetic energy
a gallob of gasoline has potential energy
rotating flywheel in a car in a car engine has kinetic energy
a swinging wrecking ball has kinetic energy
skier on a mountaintop has both

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Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

A measure of randomness or disorder

vodka [1.7K]

It is simply called Entropy.

3 0

3 years ago

Read 2 more answers

A quarterback passes a football from height h = 2.1 m above the field, with initial velocity v0 = 13.5 m/s at an angle θ = 32° a

SOVA2 [1]

Answer:

a) x = v₀² sin 2θ / g

b) t_total = 2 v₀ sin θ / g

c) x = 16.7 m

Explanation:

This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

sin θ = $v_{oy}$ / vo

cos θ = v₀ₓ / vo

v_{oy} = v_{o} sin θ

v₀ₓ = v₀ cos θ

v_{oy} = 13.5 sin 32 = 7.15 m / s

v₀ₓ = 13.5 cos 32 = 11.45 m / s

a) In the x axis there is no acceleration so the velocity is constant

v₀ₓ = x / t

x = v₀ₓ t

the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero

v_{y} = v_{oy} - gt

0 = v₀ sin θ - gt

t = v_{o} sin θ / g

we substitute

x = v₀ cos θ (2 v_{o} sin θ / g)

x = v₀² /g 2 cos θ sin θ

x = v₀² sin 2θ / g

at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,

b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

at the highest point the vertical speed is zero

v_{y} = v_{oy} - gt

v_{y} = 0

t = v_{oy} / g

t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

t_total = 2 t

t_total = 2 v₀ sin θ / g

c) we calculate

x = 13.5 2 sin (2 32) / 9.8

x = 16.7 m

5 0

3 years ago

How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?

myrzilka [38]

(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)

5 0

2 years ago

On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$