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pentagon [3]
3 years ago
5

Suppose that a solid ball, a solid disk, and a hoop all have the same mass and the same radius. Each object is set rolling witho

ut slipping up an incline with the same initial linear (translational) speed. Which object goes farthest up the incline?
Physics
1 answer:
8090 [49]3 years ago
5 0

Answer:

Hoop will reach the maximum height

Explanation:

let the mass and radius of solid ball, solid disk and hoop be m and r  (all have same radius and mass)

They all  are rolled with similar initial speed v

by the law of conservation of energy we can write

K_{trans}+K_{rot}= P

for solid ball

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{ball}\omega^2= mgh_{ball}

putting I_{ball}=\frac{2}{5}mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{ball}= 0.071v^2

now for solid disk

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{disk}\omega^2= mgh_{disk}

putting I_{ball}=\frac{1}{2}mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{disk}= 0.076v^2

for hoop

[tex]\frac{1}{2}mv^2+\frac{1}{2}I_{hoop}\omega^2= mgh_{hoop}

putting I_{hoop}=mr^2 and \omega=\frac{v}{r} in the above equation and solving we get

h_{hook}= 0.10v^2

clearly from the above calculation we can say that the Hoop will reach the maximum height

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Answer:

A compact car, with mass 725 kg, ... at 115 km/h toward the east. ... b. A second car, with a mass of 2175 kg, has the same momentum. What is its ... Glisens : m = 2175 kg;. D 21,32 xrolka. anknown. r = ? 110.6 mis east

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8 0
2 years ago
1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s
siniylev [52]

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

4 0
2 years ago
Bill is farsighted and has a near point located 121 cm from his eyes. Anne is also farsighted, but her near point is 74.0 cm fro
Arada [10]

Answer:

Explanation:

The lens equation is

1 / f = 1 / di + 1 / do

Where

f is focal length

di is the image distance

do is the object distance

Both Annie and Billy use a glass whose near point is 25cm

Then, the object distance is

do = 25 - 2 = 23cm

The have the same object distance.

Let find the vocal length of bills eye

Given that,

Bill near point is 121cm and distance of the glass from the eye is 2cm

Then,

Image distance of bill is

di_B = -(121-2) = -119cm

object distance do = 23cm

Then,

1 / f_B = 1 / di_B + 1 / do

1 / f_B = -1 / 119 + 1 / 23

1 / f_B = -119^-1 + 23^-1

1 / f_B = 0.0351

Then, f_B = 28.51 cm

Also, let find Annie focal length

Given that,

Annie near point is 74 cm and distance of the glass from the eye is 2cm

Then,

Image distance of Annie is

di_A = -(74-2) = -72cm

object distance do = 23cm

Then,

1 / f_A = 1 / di_A + 1 / do

1 / f_A = -1 / 72 + 1 / 23

1 / f_A = -72^-1 + 23^-1

1 / f_A = 0.02959

Then, f_A = 33.8 cm

Distance of object from the lens when Annie uses Billy glass

Then,

1 / f_B = 1 / di_A + 1 / do

1 / 28.51 = -1 / 72 + 1 / do

28.51^-1 = -72^-1 + do^-1

do^-1 = 28.51^-1 + 72^-1

do^-1 = 0.048964

do = 20.42 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_A = 20.42 + 2 = 22.42cm

do_A = 22.42 cm

Distance of object from the lens when Billy uses Annie glass

Then,

1 / f_A = 1 / di_B + 1 / do

1 / 33.8 = -1 / 119 + 1 / do

33.8^-1 = -119^-1 + do^-1

do^-1 = 33.8^-1 + 119^-1

do^-1 = 0.03799

do = 26.32 cm

Then, the object location from the eye will be, the eye is 2cm from the glass. Then,

do_B = 26.32 + 2 = 28.32 cm

do_B = 28.32 cm

7 0
3 years ago
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