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4 years ago

Which form of energy is involved in weighting fruit on a spring scale ?

Physics

2 answers:

galben [10]4 years ago

7 0

The answer would be ELASTIC POTENTIAL ENERGY...

Hope that helps!!!!

alexdok [17]4 years ago

5 0

The correct answer is elastic potential energy.

I hoped this helped!

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MY LAST ONE<,Brainliest for best answer!

grigory [225]

_Award brainliest if helped!
Mechanical Advantage = Force by Hammer / Force by Nail = 160/40 = 4

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3 years ago

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Can someone help me with this question

77julia77 [94]

I didnt want to type it all so here is the link that tells you all you need to answer that question

https://kids.kiddle.co/Electron_cloud

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3 years ago

How long does GSR last ?

SSSSS [86.1K]

Answer:

i think that the answer might be be C. one day. because it last about 4 to 6 hours

4 0

3 years ago

5. A 500 kg satellite is in a circular orbit at an altitude of 500 km above the Earth's surface. Because of air friction, the sa

nika2105 [10]

Answer:

the energy transformed into internal energy is E fr = 1.57*10¹⁰ J

Explanation:

From energy conservation , the internal energy gained by friction should be equal to the loss of total energy of the satellite

Since the total energy of the satellite can be decomposed into kinetic and potential energy

E fr = ΔE = E₂-E₁

E₁= K₁+V₁ = 1/2*m*v₁² + m*g*h₁

E₂ = K₂+V₂= 1/2*m*v₂² + m*g*h₂

first, we can choose our reference state so that h₂=0 and h₁=h=500 km

second , we can calculate the approximate the inicial velocity as the velocity required for a stable circular orbit

g = v₁²/(h+R) → v₁² = g*(h+R)

as the velocity diminishes, h diminishes, falling into the earth

assuming the radius of the Earth as R= 6371 km , then

v₁² = g*(h+R) = 9.8 m/s² * (500 km+ 6371 km) *1000 m/km = 6.73 * 10⁷ (m/s)²

replacing values

E₁ = 1/2*m*v₁² + m*g*h₁ = 1/2* 500kg *6.73 * 10⁷ (m/s)² + 500kg* 9.8m/s² * 500 km = 1.67*10¹⁰ J

E₂= 1/2*m*v₂² + m*g*h₂ = 1/2* 500kg *(2000 m/s)² + 0 = 1*10⁹ J

therefore

E fr = ΔE = E₂-E₁ = 1.67*10¹⁰ J - 1*10⁹ J = 1.57*10¹⁰ J

5 0

3 years ago

The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.

Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

The maximum steepness of the slope where the truck can be parked without tipping over is approximately 54.55 %.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, C represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

$tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}$

The steepness of the slope is therefore;

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100$

Where;

$\overline{AM}$ = Half the width of the truck = $\dfrac{2.4 \, m}{2}$ = 1.2 m

$\overline{CM}$ = The elevation of the center of gravity above the ground = 2.2 m

$\mathrm{The \ steepness \ of \ the \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%$

$tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}$

$Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right) \approx 28.6 ^{\circ}$

The maximum steepness of the slope where the truck can be parked is 54.55 %.

Learn more here:

brainly.com/question/20793607

3 0

3 years ago