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arsen [322]
3 years ago
8

The diagram shows a ball resting at the top of a hill.

Physics
2 answers:
andrey2020 [161]3 years ago
8 0

Answer:

The potential energy in the system is greatest at X.

The first option

WARRIOR [948]3 years ago
7 0

Answer:

a) The potential energy in the system is greatest at X.

Explanation:

Let be X the point where a ball rests at the top of a hill. By applying the Principle of Energy Conservation, the total energy in the physical system remains constant and gravitational potential energy at the top of the hill is equal to the sum of kinetic energy, a lower gravitational energy and dissipated work due to nonconservative forces (friction, dragging).

U_{grav, X} = U_{grav, Y} + K_{Y} + \Delta W_{X \rightarrow Y} = U_{grav,Z}+\Delta W_{X \rightarrow Z}

Conclusions are showed as follows:

a) The potential energy in the system is greatest at X.

b) The kinetic energy is the lowest at X and Z.

c) Total energy remains constant as the ball moves from X to Y.

Hence, the correct answer is A.

You might be interested in
(b) During one day, 250 kg of water is pumped through
ser-zykov [4K]

Answer: The energy incident on the solar panel during that day is 9.24 \times 10^{7} J.

Explanation:

Given: Mass = 250 kg

Initial temperature = 16^{o}C

Final temperature = 38^{o}C

Specific heat capacity = 4200 J/kg^{o}C

Formula used to calculate the energy is as follows.

q = m \times C \times (T_{2} - T_{1})

where,

q = heat energy

m = mass of substance

C = specific heat capacity

T_{1} = initial temperature

T_{2} = final temperature

Substitute the values into above formula as follows.

q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C

As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.

\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J

Thus, we can conclude that the energy incident on the solar panel during that day is 9.24 \times 10^{7} J.

4 0
2 years ago
Two 20.0 g ice cubes at − 20.0 ∘ C are placed into 285 g of water at 25.0 ∘ C. Assuming no energy is transferred to or from the
Lelechka [254]

Answer:

Ft = 17.48°C

Explanation:

Ft is the final temperature. However, ice absorbs heat during two process of melting and cooling and as such, there is no loss of heat to or from the surrounding hence by conservation of energy.

Therefore,

Heat absorbed by water of 20g = heat rejected by water of 265g.

So; M(ice)[C(ice) [(ΔT) + LH(ice) + C(water)(ΔT)] = C(water) M(water) (ΔT)

So, 20[(2.108) [0 - (-20)] + 333.5 + 4.187(Ft - 0)]] = (285)(4.187) (25 - Ft)

To get;

7513 + 83.74 Ft = 29832.4 - 1193.3 Ft

So factorizing, we get;

83.74 Ft + 1193.3 Ft = 29832.4 - 7513

So; 1277.04 Ft = 22319.4

So; Ft = 22319.4/1277.04 = 17.48°C

7 0
2 years ago
What is the kinetic energy of an object moving 40m/s with a mass of 20kg?
Svet_ta [14]
KE = (1/2) (mass) (speed)²

KE = (1/2) (20 kg) (40 m/s)²

KE = (1/2) (20 kg) (1,600 m²/s²)

KE = (10 kg) (1,600 m²/s²)

KE = 16,000 Joules
7 0
3 years ago
Help it’s actually physical science but plz help
Butoxors [25]
The correct answer is B
3 0
3 years ago
What is the density of a rock with a mass of .235kg
Andre45 [30]

Answer:

where is volume? formula of density is: mass/volume so volume must be there

4 0
2 years ago
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