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3 years ago

Can someone please help me I will mark you as brilliant.

Physics

1 answer:

Svet_ta [14]3 years ago

4 0

Answer:

1=8 ohms 2=0.5 Amps

Explanation:

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A student is preparing to take a bath when she realizes the hot water tap in the bathroom is not working the student go to the k

WITCHER [35]

Answer:

The final temperature of the bath water will be 55°C

Explanation:

A student is preparing to take a bath when she realizes that the hot water tap is not working.

The student goes to the kitchen and prepares 10 L of 100 degree Celsius water to mix with 10 L of 10 degree Celsius water. What will be the final temperature of the bath water?

Using the relation, Heat lost = Heat gained (assuming no heat is lost to the surroundings).

Heat lost by hot water = heat gained by cold water

Using the formula of heat, Q = mcΔT

where m = mass of water, c = specific heat capacity of water, ΔT = temperature difference

mass of water = density * volume (density of water = 1 kg/L)

mass of hot water, m₁ = 1 kg/L * 10 L = 10 kg;

mass of cold water, m₂ = 1 kg/L * 10 L = 10 kg;

specific heat capacity of water, c = 4200 j/kg

Let the final temperature of the mixture be T

-m₁cΔT₁ = m₂cΔT₂ ( since heat is lost by the hot water)

-m₁ΔT₁ = m₂ΔT₂

-10 * (T- 100) = 10 * (T - 10)

-10T + 1000 = 10T - 100

20T = 1100

T = 1100/20

T = 55°C

Therefore, the final temperature of the bath water will be 55°C

6 0

3 years ago

It is weigh-in time for the local under 85 kg rugby team. The bathroom scale used to assess eligibility can be described by Hook

Grace [21]

15 m/n is the answer

4 0

3 years ago

A 75.0 W bulb is connected to a

adoni [48]

Answer:

Explanation:

W = 75 watts

V = 110 volts

Formula

W = V * I

Solution

75 = 110 * I Divide by 110

75 / 110 = I

I = 0.6818 Amperes

6 0

3 years ago

New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in

seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

$v = \lambda * f (1)$

Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

$\lambda_{low} = \frac{c}{f_{high}} = \frac{3e8m/s}{300e9Hz} = 1 mm (2)$

$\lambda_{high} = \frac{c}{f_{low}} = \frac{3e8m/s}{30e9Hz} = 10 mm (3)$

4 0

3 years ago

How would you describe the movement of a transverse wave?

nlexa [21]

Answer:

Up AND Down

Explanation:

a transverse wave goes up and Down

4 0

3 years ago