Relative to the distance of an object in front of a plane mirror, how far behind the mirror is the image?

1. Consider three objects: Object A is a hoop of mass m and radius r; Object B is a sphere

Zarrin [17]

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Explanation:

The moments of inertia of the three objects are the following:

1) For a hoop of negligible thickness, it is

$I=MR^2$

where M is its mass and R its radius. For the hoop in this problem,

M = m

R = r

Therefore, its moment of inertia is

$I=(m)(r)^2=mr^2$

2) For a solid sphere, the moment of inertia is

$I=\frac{2}{5}MR^2$

where M is its mass and R its radius. For the sphere in this problem,

M = 2m

R = r

Therefore, its moment of inertia is

$I=\frac{2}{5}(2m)(r)^2=\frac{4}{5}mr^2$

3) For a disk of negligible thickness, the moment of inertia is

$I=\frac{1}{2}MR^2$

where M is its mass and R its radius. For the disk in this problem,

M = 3m

R = r

Therefore, its moment of inertia is

$I=\frac{1}{2}(3m)(r)^2=\frac{3}{2}mr^2$

So now we can answer the two questions:

a) Object C (the disk) has the greatest rotational inertia ( $\frac{3}{2}mr^2$ )

b) Object B (the sphere) has the smallest rotational inertia ( $\frac{4}{5}mr^2$ )

Learn more about inertia:

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7 0

4 years ago

Velocity of a Hot-Air Balloon A hot-air balloon rises vertically from the ground so that its height after t sec is given by the

Inessa [10]

Answer:

Explanation:

Given the height reached by a balloon after t sec modeled by the equation

h=1/2t²+1/2t

a) To calculate the height of the balloon after 40 secs we will substitute t = 40 into the modeled equation and calculate the value of t

If h(t)=1/2t²+1/2t

h(40) = 1/2(40)²+1/2 (40)

h(40) = 1600/2 + 40/2

h(40) = 800 + 20

h(40) = 820 feet

The height of the balloon after 40 secs is 820 feet

b) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

when v = 0sec

v(0) = 0 + 1/2

v(0) = 1/2 ft/sec

at v = 30secs

v(30) = 30 + 1/2

v(30) = 30 1/2 ft/sec

average velocity = v(30) - v(0)

average velocity = 30 1/2 - 1/2

average velocity of the balloon between t = 0 and t = 30 = 30 ft/sec

c) Velocity is the change of displacement of a body with respect to time.

v = dh/dt

v(t) = 2(1/2)t²⁻¹ + 1/2

v(t) = t + 1/2

The velocity of the balloon after 30secs will be;

v(30) = 30+1/2

v(30) = 30.5ft/sec

The velocity of the balloon after 30 secs is 30.5 feet/sec

6 0

4 years ago

Please Help Me with This

densk [106]

Answer:

Explanation:

a). Find the graph attached for the motion.

b). If a shopper walk 5.4 m westwards then 7.8 m eastwards,

Distance traveled by the shopper = Distance traveled in eastwards + Distance traveled westwards

= 5.4 + 7.8

= 13.2 m

c). Displacement of the shopper = Distance walked westwards - Distance traveled eastwards

= 5.4 - 7.8

= -2.4 m

Therefore, magnitude of the displacement of the shopper is = 2.4 m

And the direction of the displacement is eastwards.

8 0

3 years ago

Anyone who is willing to help me do some 8th grade science questions<br>NO LINKS

NISA [10]

Answer:

what are they ill have a look

Explanation:

3 0

3 years ago

The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How

qwelly [4]

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

7 0

3 years ago

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