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Angelina_Jolie [31]
2 years ago
13

1. The spring in a trampoline has been stretched 0.25m and has a spring

Physics
1 answer:
nata0808 [166]2 years ago
4 0

Answer:

250/0.25. * 0.25 =250J

Explanation:

PE=Force*displacement

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(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
Directions: Fill in the blank with the correct word. Choose from the list of possible answers in the word bank below:
Alexxx [7]

Answer:

1).atoms (3). mixture. (5). Element

2). particles (4). molecules (6). suspension

Explanation:

(7). Homogeneous (8). Heterogeneous

(9). compound (10). solutions

8 0
3 years ago
Which of the following materials is necessary to stop an alpha particle? a. three feet of concrete c. single sheet of aluminum f
Nat2105 [25]
A single sheet of paper can stop an alpha particle
8 0
3 years ago
Read 2 more answers
A horse of mass 242 kg pulls a cart of mass 224 kg. The acceleration of gravity is 9.8 m/s 2 . What is the largest acceleration
Rufina [12.5K]

To solve this problem it is necessary to apply the concepts related to Newton's second Law and the force of friction. According to Newton, the Force is defined as

F = ma

Where,

m= Mass

a = Acceleration

At the same time the frictional force can be defined as,

F_f = \mu N

Where,

\mu = Frictional coefficient

N = Normal force (mass*gravity)

Our values are given as,

m_h = 242 kg\\m_c = 224 kg\\\mu = 0.894\\

By condition of Balance the friction force must be equal to the total net force, that is to say

F_{net} = F_f

m_{total}a = \mu m_hg

(m_h+m_c)a = \mu*m_h*g

Re-arrange to find acceleration,

a= \frac{\mu*m_h*g}{(m_h+m_c)}

a = \frac{0.894*242*9.8}{(242+224)}

a = 4.54 m/s^2

Therefore the acceleration the horse can give is 4.54m/s^2

3 0
3 years ago
A 6.0-kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
jeka57 [31]

Answer:

a. 3.1 m/s^2

Explanation:

The equation of the forces along the directions parallel and perpendicular to the slope are:

- Along the parallel direction:

mg sin \theta - \mu_k R = ma

where :

m = 6.0 kg is the mass  of the box

g = 9.8 m/s^2 the acceleration of gravity

\theta=39^{\circ}  is the angle of the slope

\mu_k = 0.40 is the coefficient of friction

R is the normal reaction  

a is the acceleration

- Along the perpendicular direction:

R-mg cos \theta =0

From the 2nd equation, we get an expression for the reaction force:

R=mg cos \theta

And substituting into the 1st equation, we can find the acceleration:

mg sin \theta - \mu_k mg cos \theta = ma

Solving for a,

a=g sin \theta - \mu_k g cos \theta =(9.8)(sin 39^{\circ})-(0.40)(9.8)(cos 39^{\circ})=3.1 m/s^2

6 0
3 years ago
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