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sineoko [7]
3 years ago
11

Where is a trench most likely to occur?

Physics
2 answers:
Dafna1 [17]3 years ago
7 0
D. convergent plate boundary involving an oceanic plate
Dafna11 [192]3 years ago
3 0
The answer is d convergent plate boundary involving an oceanic plate
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The force of gravity acting on an object is the object's ______. A. acceleration B. mass C. weight D. matter
ryzh [129]

Answer:

d

Explanation:

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2 years ago
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Select the three words in the following sentence that indicate it is in the third person.
Triss [41]
Die meisten von ihnen haben die Möglichkeit zu den Anderen zu kommen oder die Möglichkeit für die Zeit der Arbeit mit dem Auto und der Wohnung zu
4 0
2 years ago
D=1/2at^2 <br> solve for a
nignag [31]

Answer:

a = 2d / t²

Explanation:

d = ½ at²

Multiply both sides by 2:

2d = at²

Divide both sides by t²:

a = 2d / t²

4 0
3 years ago
In which of the two situations described is more energy transferred?
Furkat [3]

Answer:

More energy is transferred in situation A

Explanation:

Each of the situations are analyzed as follows;

Situation A

The temperature of the cup of hot chocolate = 40 °C

The temperature of the interior of the freezer in which the chocolate is placed = -20 °C

We note that at 0°C, the water in the chocolate freezes

The energy transferred by the chocolate to the freezer before freezing is given approximately as follows;

E₁ = m×c₁×ΔT₁

Where;

m = The mass of the chocolate

c₁ = The specific heat capacity of water = 4.184 kJ/(kg·K)

ΔT₁ = The change in temperature from 40 °C to 0°C

Therefore, we have;

E₁ = m×4.184×(40 - 0) = 167.360·m kJ

The heat the coffee gives to turn to ice is given as follows;

E₂ = m·H_f

Where;

H_f = The latent heat of fusion = 334 kJ/kg

∴ E₂ = m × 334 kJ/kg = 334·m kJ

The heat required to cool the frozen ice to -20 °C is given as follows;

E₃ = m·c₂·ΔT₂

Where;

c₂ = The specific heat capacity of ice = 2.108 kJ/(kg·K)

Therefore, we have;

E₃ = m × 2.108 ×(0 - (-20)) = 42.16

E₃ = 42.16·m kJ/(kg·K)

The total heat transferred = (167.360 + 334 + 42.16)·m kJ/(kg·K) = 543.52·m kJ/(kg·K)

Situation B

The temperature of the cup of hot chocolate = 90 °C

The temperature of the room in which the chocolate is placed = 25 °C

The heat transferred by the hot cup of coffee, E, is given as follows;

E = m×4.184×(90 - 25) = 271.96

∴ E = 271.96 kJ/(kg·K)

Therefore, the total heat transferred in situation A is approximately twice the heat transferred in situation B and is therefore more than the heat transferred in situation B

Energy transferred in situation A = 543.52 kJ/(kg·K)

Energy transferred in situation B = 271.96 kJ/(kg·K)

Energy transferred in situation A ≈ 2 × Energy transferred in situation B

∴ Energy transferred in situation A > Energy transferred in situation B.

3 0
3 years ago
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
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