Answer:
Ratio of series current to parallel
= 1 : 8
Explanation:
Total resistance Rt
For series, Rt = 2+2+2+2 = 4ohms
For parallel, 1/Rt = 1/2 + 1/2 + 1/2 + 1/2
1/Rt = 4/2, Rt = 2/4 ohms.
If we use a 1V battery, then,
I = V/Rt
I = 1/4 = 0.25 ampere for series arrangement.
I = 1/0.5 = 2 ohms.
Ratio of current of series to parallel = 0.25 : 2
= 1 : 8
He would try to enter as the tide is rising, and leave as the tide is falling. Those things happen at all different times of day during a month.
Answer:
PE = 44.1 J
Explanation:
Ok, to have the specific data, the first thing we must do is convert from grams to kilograms. Since mass must always be in kilograms (kg)
We have:
- 1 kilograms = 1000 grams.
We convert it using a rule of 3, replacing, simplifying units and solving:
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Earth's gravity is known to be 9.8 m/s², so we have:
Data:
- m = 0.3 kg
- g = 9.8 m/s²
- h = 15 m
- PE = ?
Use formula of potencial energy:
Replace and solve:
Since the decimal number, that is, the number after the comma is less than 5, it cannot be rounded, then we have this result.
The potential energy of the volleyball is <u>44.1 Joules.</u>
Greetings.
Answer: C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.
Explanation:
The speed of each car is defined as:

where d is the distance traveled by the car and t is the time taken.
For the yellow car, d=400 mi and t=8 h, so its speed is

For the green car, d=400 mi and t=10 h, so its speed is

So, the correct choice is
C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.