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3 years ago

Instructions:Select all the correct answers.

2 answers:

3 0

Choice (3) would be one of the objects because when its stretching its waiting to release the energy. Then choice (1) would be the other object because when it rolls the energy stored in the ball would be released when stopped. Therefore there storing the energy until released.

V125BC [204]3 years ago

3 0

Answer:

2. A small rock sitting on the top of big rock.

3. a stretched rubber band.

Explanation:

The potential energy is defined as the energy by the virtue of the position of the object, its electric energy, stress which present in itself.

There are the two forms of potential energies which are more common:

Gravitational potential energy: It is a form of potential energy which depends on the mass of an object and the distance of that object from the center of the earth. In the given situation the rock which is sitting on the top of big rock posses gravitational potential energy.
Elastic potential energy: It is the form of energy which is produced by the virtue of extending spring. This energy in this is the store energy of the extending or compressed spring. In the given situation a stretched rubber band is the example of elastic potential energy.

Therefore, in the given problem a stretched rubber band and a small rock which is sitting on the top of big rock have stored energy.

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The combined-gas law relates which of these?

Fofino [41]

The combined-gas law relates which temperature, pressure and volume.

Temperature=T
Pressure=P
Volume=V

(P₁*V₁) / T₁=(P₂*V₂) / T₂

D. Temperature, pressuere and volume.

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3 years ago

A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.

Luba_88 [7]

Answer:

The final equilibrium temperature of the system is $T = 12.48^oC$

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice $M_{i}$ = 129 g = 0.129 kg

Mass of the steam $M_s$ = 19 g = 0.019 kg

Initial temperature is $T_i$ = 0°C

Temperature of steam $T_s$ = 100°C

Following the change of state of water in the question

The energy required by ice to change to water is mathematically given as

$Q_A = M_iL_f$

Where $L_f$ is a constant known as heat of fusion and the value is $334*10^3 J/kg$

$Q_A = 0.129 *334 *10^3 = 43086 J$

The energy been released when the steam changes to water is mathematically given as

$Q_B = M_s * L_v$

Where $L_v$ is a constant known as heat of vaporization and the value is $2256*10^3J/kg$

$Q_B = 0.019 * 2256*10^3 = 42864J$

The energy released when the temperature of water decrease from 100°C to 0°C is

$Q_C = M_s *C_water ($ 100°C)

Where $C_{water}$ is the specific heat of water which has a value $4186J/kg \cdot K$

$Q_C = 0.019 *4186*100 = 7953.4$

Looking at the values we obtained we noticed that ]

$Q_B + Q_C > Q_A$

What this means is that the ice will melt

bearing in mind the conservation of energy

looking the way at which water at different temperature were mixed according to the question

Heat lossed by the vapor = heat gained by ice

$Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T$

$T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}$

$T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}$

$T = 12.48^oC$

3 0

4 years ago

For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas

nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

$F=\dfrac{mv^2}{R}$ , m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

$F=\dfrac{(16)^2m}{60}=4.26m\ N$

Case 2.

R = 15 m v = 8 m/s

$F=\dfrac{(8)^2m}{15}=4.26m\ N$

Case 3.

R = 30 m v = 4 m/s

$F=\dfrac{(4)^2m}{30}=0.54m\ N$

Case 4.

R = 45 m v = 4 m/s

$F=\dfrac{(4)^2m}{45}=0.36m\ N$

Case 5.

R = 30 m v = 16 m/s

$F=\dfrac{(16)^2m}{30}=8.54m\ N$

Case 6.

R = 15 m v =12 m/s

$F=\dfrac{(12)^2m}{15}=9.6m\ N$

Ranking from largest to smallest is given by :

F>E>A=B>C>D

5 0

3 years ago

A force is applied to a 30-kg object, resulting in an acceleration of 15 m/s. What is the force being applied to the object?

telo118 [61]

The sum of all forces applied to the object = (mass) x (acceleration)

The sum of all forces applied to the object = (30) x (15)

= <u>450 newtons</u>, in the direction of the acceleration.

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4 years ago

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A temperature reading of zero on the Celsius scale is equal to absolute zero.

GrogVix [38]

No, that's false. 'Zero' on the Celsius scale
is 273 on the absolute scale.

8 0

4 years ago

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