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Viefleur [7K]
2 years ago
14

Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal

line, Group of answer choices the kinetic energy of the block when it has fully compressed the spring will be the kinetic energy of the block just before it collides with the spring will be equal to mgh. the kinetic energy of the block when it has fully compressed the spring will be equal to mgh. the kinetic energy of the block when it has fully compressed the spring will be zero. the kinetic energy of the block just before it collides with the spring will be kx2. Not saved Questions AnsweredQuestion 1 AnsweredQuestion 2 AnsweredQuestion 3 AnsweredQuestion 4 AnsweredQuestion 5 AnsweredQuestion 6 AnsweredQuestion 7 AnsweredQuestion 8 AnsweredQuestion 9 Haven't Answered YetQuestion 10 Time Elapsed: Hide Attempt due: Oct 12 at 11:59pm
Physics
1 answer:
Leya [2.2K]2 years ago
4 0

The energy in the system is given by the law of conservation of energy.

The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h.

Reason:

The given parameters are;

Surface of the inclined plane = Frictionless

Potential energy at the horizontal line = Zero of gravitational potential

By the law of Conservation of Energy, we have;

The spring will be compressed by a distance, <em>x</em>

The energy stored in the compressed spring, K.E. = (1/2)·k·x²

Energy in the block when the block comes to rest at a height, h₁, will be, P.E. = m·g·h₁

Therefore, by conservation of energy, we have;

The initial potential of the block = The stored energy in the compressed string + The gravitational potential energy of the block when it has compressed.

Therefore, the correct option is; <u>The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h</u>

Learn more here;

brainly.com/question/17713698

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The international space station travels at a distance of about 250 miles above Earth’s surface and at a speed of 17,500 miles pe
Arisa [49]

Answer:

In this case we are dealing with the pythagorean theorm involving right angled triangles. This theorm states that a^2 + b^2 = c^2 which means the square of the hypotenuse (side c, opposite the right angle) is equal to the square of the remaining two sides.

In this case we will say that a = 3963 miles which is the radius of the earth. c is equal to the radius of the earth plus the additional altitude of the space station which is 250 miles; therefore, c = 4213 miles. We must now solve for the value b which is equal to how far an astronaut can see to the horizon.

(3963)^2 + b^2 = (4213)^2

b^2 = 2,044,000

b = 1430 miles.

The astronaut can see 1430 miles to the horizon.

Explanation:

:D hopes this Helps

3 0
3 years ago
A heat engine has three quarters the thermal efficiency of a carnot engine operating between temperatures of 65°C and 435°C if
const2013 [10]

Answer:

The value is  P_e  =  31275.2 \  W

Explanation:

From the question we are told that

   The efficiency of the carnot engine is  \eta

    The efficiency of a heat engine is k =  \frac{3}{4}  *  \eta

    The operating temperatures of the carnot engine is  T_1  =  65 ^oC =338 \ K  to  T_2 = 435  ^oC = 708 \  K

    The rate at which the heat engine absorbs energy is  P =  44.0 kW  = 44.0 *10^{3} \  W

Generally the efficiency of the carnot engine is mathematically represented as

          \eta =  [ 1 - \frac{T_1 }{T_2}  ]

=>       \eta =  [ \frac{T_2 - T_1}{T_2} ]

=>       \eta = 0.3856

Generally the efficiency of the heat engine is

           k  =  \frac{3}{4} * 0.3856

=>        k  = 0.2892

Generally the efficiency of the heat engine is also mathematically represented as

          k  =  \frac{W}{P}

Here W is the work done which is mathematically represented as

        W =  P - P_e

Here P_e  is the heat exhausted

So

       k  =  \frac{P - P_e}{P}

=>    0.2892   =  \frac{44*10^{3} - P_e}{44*10^{3}}

=>   P_e  =  31275.2 \  W

8 0
3 years ago
Determine the force of gravitational attraction between the Earth and the moon. Their masses are 5.98 x 1024 kg and 7.26 x 1022
monitta

Answer:

F=1.95\times 10^{20}\ N

Explanation:

Mass of Earth, m_e=5.98 \times 10^{24}\ kg

Mass of Moon, m_m=7.26\times 10^{22}\ kg

The distance between Earth and the Moon is, d=384,400\ km

We need to find the force of gravitational attraction between the Earth and the moon. The force of gravity is given by :

F=G\dfrac{m_em_m}{r^2}\\\\F=6.67\times 10^{-11}\times \dfrac{5.98 \times 10^{24}\times 7.26\times 10^{22}}{(384400 \times 10^3)^2}\\\\F=1.95\times 10^{20}\ N

So, the required force is 1.95\times 10^{20}\ N.

4 0
3 years ago
What is the final concentration of DD at equilibrium if the initial concentrations are [A][A]A_i = 1.00 MM and [B][B]B_i = 2.00
pentagon [3]

Answer:

A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.

B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.

[D] = 0.90 M

Explanation:

With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.

This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.

- The first attached image contains the correct question.

- The solution to part A is presented in the second attached image.

- The solution to part B is presented in the third attached image.

8 0
3 years ago
A ship, carrying freshwater to a desert island in the caribbean, has a horizontal cross-sectional area of 2800 m2 at the waterli
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3 0
3 years ago
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