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Korvikt [17]
3 years ago
10

Which object will have the most potential energy?

Physics
1 answer:
Svetradugi [14.3K]3 years ago
5 0

Answer:  the most potential energy ==  5 kg book, 2 m from the ground= 98 Joules

Explanation:

potential energy = m g h

m = mass

g = acceleration due gravity  = 9.8 m/s²

h = distance above  ground

1.  Pe₁ = 1 kg x 2 m x g  = 2 g

2. Pe₂ = 5 kg x 2 m x g = 10 g = 10 kg m x 9,8 m/s² = 98 Joules

3. Pe₃ = 1 kg x 0,5 m x g = 0,5 g

4. Pe₄ = 5 kg x 0.5 m x g = 2,5 g  

10 > 2,5 > 2 >0,5

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A 10 kg mass car initially at rest on a horizontal track is pushed by a horizontal force of 10 N magnitude. If we neglect the fr
vlada-n [284]

Answer:

50 m

Explanation:

F = ma

10 N = (10 kg) a

a = 1 m/s²

Given:

v₀ = 0 m/s

a = 1 m/s²

t = 10 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (0 m/s) (10 s) + ½ (1 m/s²) (10 s)²

Δx = 50 m

5 0
3 years ago
Dr. Kirwan is preparing a slide show that he will present to the executive board at tonight's committee meeting. He places a 3.5
lisov135 [29]

Answer:

A) d_o = 20.7 cm

B) h_i = 1.014 m

Explanation:

A) To solve this, we will use the lens equation formula;

1/f = 1/d_o + 1/d_i

Where;

f is focal Length = 20 cm = 0.2

d_o is object distance

d_i is image distance = 6m

1/0.2 = 1/d_o + 1/6

1/d_o = 1/0.2 - 1/6

1/d_o = 4.8333

d_o = 1/4.8333

d_o = 0.207 m

d_o = 20.7 cm

B) to solve this, we will use the magnification equation;

M = h_i/h_o = d_i/d_o

Where;

h_o = 3.5 cm = 0.035 m

d_i = 6 m

d_o = 20.7 cm = 0.207 m

Thus;

h_i = (6/0.207) × 0.035

h_i = 1.014 m

8 0
3 years ago
A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
A large body collides with a stationary small body from the rear end. How do the bodies behave if the collision is inelastic?.
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If the collision is inelastic, there is every possibility that the large body will drag the small stationary body along with it in the direction of the collision. Some amount of heat, light and sound energy will also be produced due to the kinetic energy of the large body. I hope the answer helps you.
6 0
3 years ago
How do we represent Earth’s face?
jeka94

Answer:

Human-driven changes in arrive utilize and arrive cover such as deforestation, urbanization, and shifts in vegetation designs moreover change the climate, coming about in changes to the reflectivity of the Soil surface (albedo), emanations from burning timberlands, urban warm island impacts and changes within the normal water cycle.

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