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damaskus [11]
2 years ago
14

Dane is standing on the moon holding an 8 kilogram brick 2 metres above the ground. How much energy is in the brick's gravitatio

nal potential energy store? (On the moon the gravitational field strength is 1.6 N/kg). Give your answer to 1 d.p.​
Physics
2 answers:
PSYCHO15rus [73]2 years ago
6 0

Answer:

25.6 Joules

Explanation:

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

Read more on Brainly.com - brainly.com/question/14383332#readmore

Nadya [2.5K]2 years ago
3 0

The gravitational potential energy of the brick is 25.6 J

Explanation:

The gravitational potential energy of an object is the energy possessed by the object due to its position in a gravitational field.

Near the surface of a planet, the gravitational potential energy is given by

PE=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the height of the object relative to the ground

For the brick in this problem, we have:

m = 8 kg is its mass

g = 1.6 N/kg is the strenght of the gravitational field on the moon

h = 2 m is the height above the ground

Substituting, we find:

PE=(8)(1.6)(2)=25.6 J

Learn more about potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

#LearnwithBrainly

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What is the mass of an object weighing 63 N on Earth?
avanturin [10]
Weight expressed in Newtons is expressed in the equation whereby Weight= the mass of an object * the force of gravity. The force of gravity on earth is a constant 9.8 meters per second squared. Therefore if weight (w) = 63 N and the force of gravity is 63 N then the mass must equal 6.43 kg. Because the equation for weight is w=mg so 63 N (w) = m * 9.8 m/s^2. 
3 0
2 years ago
2. If a cyclist in the Tour de France traveled southwest a distance of 12,250 meters in one hour, what would the velocity of the
Luda [366]
  • Answer:

<em>12,25 km/h</em>

<em>≈ 3,4 m/s </em>

  • Explanation:

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>= 12,25km/h</em>

<em>or</em>

<em>v = d/t</em>

<em>= 12250m/h</em>

<em>1h = 60m×60s = 3600s</em>

<em>= 12250m/3600s</em>

<em>≈ 3,4 m/s </em>

5 0
2 years ago
Use the periodic table and your knowledge of isotopes to complete these statements.
klasskru [66]
<span><span>Use the periodic table and your knowledge of isotopes to complete these statements.

When polonium-210 emits an alpha particle, the child isotope has an atomic mass of </span><span> ⇒ 206</span>.</span>

<span><span>I-131 undergoes beta-minus decay. The chemical symbol for the new element is </span><span> ⇒ Xe</span>.</span>

<span><span>Fluorine-18 undergoes beta-plus decay. The child isotope has an atomic mass of </span><span> ⇒ 18</span>.</span>

7 0
2 years ago
Read 2 more answers
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
2 years ago
hockey puck slides across the ice with an initial velocity of 7.2 m/s. It has a deceleration of 1.1 m/s2 and is traveling toward
dimulka [17.4K]

For this use the formula:

d = Vo * t - (at^2) / 2

Clearing t:

t = d/(v + 0.5*a)

Replacing:

t = 5 m / (7.2 m/s + 0.5 * (-1.1 m/s²)

Resolving:

t = 5 m / (7.2 m/s + (-0.55 m/s²)

t = 5 m / 6.65 m/s

t = 0.75 s

Result:

The time will be <u>0.75 seconds.</u>

7 0
2 years ago
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