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Alex Ar [27]
3 years ago
5

Metric system please help me I don’t know how to do this

Physics
1 answer:
amm18123 years ago
8 0

Answer:

11. 6700L

12. 0.0000238

13. 1805

14. 3800

15. 1400

16. 0.01428

17. 0.0003585

18. 0.0415

19. 4060.0

20.0.105

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A 300W hot plate produces 45,000 J of thermal energy while operating for 2 min. What is the efficiency of this devide? it need t
Gekata [30.6K]

Answer:

Explanation: P = 300 W   and t = 2 min = 120 s

Energy Q = Pt = 300 W · 120 s = 36 000 J.

Thus, plate can not produce 45 000 J heat.

5 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
A charged particle moves through a magnetic field. In which situation is the magnetic force zero?
maksim [4K]

Answer:

The answer is the option a.

Explanation:

We know that magnetic force (Fm) is defined as

Fm = q (v x B)

Where q is a the value of the charge, v is the velocity of the charge and B is the value of the magnetic field.

"v x B" is defined as the cross product between the vectors velocity and magnetic field, and if the angle between them is thetha < 180°, then, the cross product is

v x B = vBsin (thetha)

So,

Fm = qvBsin (thetha)

And, in case in which v and B are parallel vectors, thetha is zero, and,

sin (thetha)=sin (0) = 0

So, Fm=0

7 0
3 years ago
Velocity: You leave on a 549-mi trip in order to attend a meeting that will start 10.8 h after you begin your trip. Along the wa
rewona [7]

Answer:

1.55 h

Explanation:

Let the time to spend over dinner be t.

Since I need to spend 10.8 h for the whole trip, then I have 10.8 - t hours left to drive.

Speed = distance/time

v = \dfrac{d}{10.8-t}

d = v(10.8-t)

The distance is 549 and maximum speed is 65 mi/h.

549 = 65(10.8 - t)

10.8-t = 8.45

t = 10.8 - 8.45 = 1.55 \text{ h}

5 0
3 years ago
When the Sun is directly overhead, a hawk dives toward the ground with a constant velocity of 5 m/s at 60 degrees below the hori
Gemiola [76]
The shadow that will be protected on the ground will represent the hawk's horizontal position. therefore it will move at the hawk's horizontal velocity component, which is:
5*cos(60) = 2.5 m/s
6 0
3 years ago
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