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3 years ago

A baseball is hit into the air with a bat. When does the baseball have the greatest gravitational potential energy?

Physics

2 answers:

fiasKO [112]3 years ago

7 0

Answer:

Explanation:

When it reaches the highest point

Klio2033 [76]3 years ago

3 0

Gravitational potential energy is related to an object's height. The greater the height, the greater the gravitational potential energy. Therefore, the baseball has the greatest gravitational potential energy when it is at its maximum height in the air.

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A vaulter holds a 26.90-N pole in equilibrium by exerting an upward force U with her leading hand and a downward force D with he

ivolga24 [154]

Answer:

Check attachment for better understanding

Explanation:

Given that

a= 0.75m

b=1.31m

c= 2.2m

Weight of pole is 26.90

Then, Fg = Weight = 26.90

Using Equilibrium of forces

ΣFy = 0

U — D — Fg = 0

U — D = Fg

U — D = 26.9

To calculate U,

We will take moment about point A.

ΣMa = 0

Let the clockwise moment be positive and anti-clockwise be negative

Fg(a+b) — U(a) = 0

26.9(1.31+0.75) —0.75U = 0

26.9(2.06) = 0.75U

0.75U = 55.414

U = 55.414/0.75

U = 73.89 N

To calculate D,

U — D = 26.9

73.89—D =26.9

73.89—26.9 = D

D = 46.99N

3 0

3 years ago

Read 2 more answers

1. If an object that stands 3 centimeters high is placed 12 centimeters in front of a plane

igor_vitrenko [27]

Answer:

1. 12 cm

2. 0.133 m

3. 0.03 m

4. Plane mirror

Virtual image

Upright

Behind the mirror

The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Real image

Inverted image

In front of the the mirror

Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Virtual image

Upright image

Behind the the mirror

Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object

Explanation:

1. For plane mirror, since there is no magnification, the virtual image distance from the mirror = object distance from the mirror = 12 cm behind the mirror

2. The height of the object = 0.3 m

The distance of the object from the mirror = 0.4 meters

Height of image formed = 0.1 meter

We have;

$Magnification, \ m = \dfrac{Image \ height }{Object \ height } = \dfrac{Image \ distance \ from \ mirror }{Object\ distance \ from \ mirror }$

$m = \dfrac{0.1}{0.3 } = \dfrac{Image \ distance \ from \ mirror }{0.4 }$

Image distance from the mirror = 0.1/0.3×0.4 = 2/15 = 0.133 m

Image distance from the mirror = 0.133 m

3. $m = \dfrac{Image \ height}{0.10 } = \dfrac{0.06 }{0.20 }$

The image height = 0.06/0.2×0.1 = 3/100 = 0.03 meter

The image height = 0.03 meter

4. Plane mirror

Type = Virtual image

Appearance = Upright image with the left transformed to right

Placement = Behind the mirror

Size = The same size as the object

Concave mirror when the object is located a distance greater than the focal length from the mirror's surface

Type = Real image

Appearance = Inverted image

Placement = In front of the the mirror

Size = Diminished when the object is beyond the center of curvature

Same size as object when the object is placed at the center of curvature

Enlarged when the object is placed between the center of curvature of the mirror and the focus of the mirror

Concave mirror when the object is located a distance less than the focal length from the mirror's surface

Type = Virtual image

Appearance = Upright image

Placement = Behind the the mirror

Size = Enlarged

Convex mirror

Type = Virtual image

Appearance = Upright image

Placement = Behind the mirror

Size = Smaller than the object.

3 0

4 years ago

A light beam in glass (n = 1.5) reaches an air-glass interface, at an angle of 60 degrees from the surface. What is the angle of

tester [92]

Answer:

θ₂ = 35.26°

Explanation:

given,

refractive index of air, n₁ = 1

refractive index of glass, n₂ = 1.5

angle of incidence, θ₁ = 60°

angle of refracted light, θ₂ = ?

using Snell's Law

n₁ sin θ₁ = n₂ sin θ₂

1 x sin 60° = 1.5 sin θ₂

sin θ₂ = 0.577

θ₂ = sin⁻¹(0.577)

θ₂ = 35.26°

Hence, the refracted light is equal to θ₂ = 35.26°

7 0

3 years ago

What important attitudes do successful scientists possess?

olganol [36]

They would use the Scientific method

8 0

3 years ago

Read 2 more answers

The _____ of a wave is the number of wave cycles per second. this quantity is given the symbol ν and has units of s–1 or _____.

Fynjy0 [20]

The frequency of a wave is the number of wave cycles per second. this quantity is given the symbol ν and has units of s^–1 or hertz.

What is a wave ?

A wave is a dynamic disturbance of one or more quantities that propagates through time. When waves oscillate frequently around an equilibrium value at a certain frequency, they are said to be periodic.

What is a wave cycle ?

The wavelength is the measured distance in meters between two waves' peaks or troughs, which is referred to as a wave cycle. It's crucial to remember that the distance from peak to peak equals the distance from trough to trough. The length of time it takes a wave to complete one cycle is known as the wave's period.

Learn more about waves here:-

brainly.com/question/3639648

#SPJ4

8 0

1 year ago