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patriot [66]
3 years ago
5

(eopt)s. Find the power dissipated in a 25 r heaterconnected to a 220V outer.​

Physics
1 answer:
Alex Ar [27]3 years ago
6 0

Answer:

P = 1936 W

Explanation:

Given that,

Voltage, V = 220 V

The resistance of the heater, R = 25 ohms

We need to find the power dissipated in the heater. We know that the power dissipated by the electric appliances is given by :

P=\dfrac{V^2}{R}\\\\P=\dfrac{(220)^2}{25}\\\\P=1936\ W

So, the required power is equal to 1936 W.

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Define a combination circuit.
Neporo4naja [7]

Answer:

a circuit that is a blend of series paths and parallel paths. See Figure for a visual explanation. Most circuits are of this variety. Don't be afraid to tackle these circuits as far as the math goes.

7 0
2 years ago
A group of animals have short, narrow legs that are partly webbed. In which environment do these animals MOST LIKELY live? A) de
Burka [1]
I'd say it's answer D) rainforests
7 0
3 years ago
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A motorcycle is traveling up one side of a hill and down the other side. The crest of the hill is a circular arc with a radius o
Setler79 [48]

Answer:

21.8 m/s

Explanation:

At the top of the hill (crest), there are two forces acting on the motorcycle:

- The reaction force of the road, N (upward)

- The force of gravity, mg (downward)

Since the motorcycle is moving by circular motion, the resultant of these forces will give the centripetal force, so:

mg-N = m\frac{v^2}{r}

where the direction of the weight (mg) is equal to that of the centripetal force, and where

m is the mass of the cycle

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed

r = 48.6 is the radius of the hill

The cycle loses contact with the road when the reaction force becomes zero:

N = 0

Substituting into the equation, we therefore find the maximum speed that is allowed for the cycle before losing constact:

mg = m\frac{v^2}{r}\\v=\sqrt{gr}=\sqrt{(9.8)(48.6)}=21.8 m/s

6 0
2 years ago
A well lagged copper calorimeter of mas 120g contains 70g of water and 10g ice both at 0°C . Dry steam at 100°C is passed in unt
Lina20 [59]

Answer:

7.6 g

Explanation:

"Well lagged" means insulated, so there's no heat transfer between the calorimeter and the surroundings.

The heat gained by the copper, water, and ice = the heat lost by the steam

Heat gained by the copper:

q = mCΔT

q = (120 g) (0.40 J/g/K) (40°C − 0°C)

q = 1920 J

Heat gained by the water:

q = mCΔT

q = (70 g) (4.2 J/g/K) (40°C − 0°C)

q = 11760 J

Heat gained by the ice:

q = mL + mCΔT

q = (10 g) (320 J/g) + (10 g) (4.2 J/g/K) (40°C − 0°C)

q = 4880 J

Heat lost by the steam:

q = mL + mCΔT

q = m (2200 J/g) + m (4.2 J/g/K) (100°C − 40°C)

q = 2452 J/g m

Plugging the values into the equation:

1920 J + 11760 J + 4880 J = 2452 J/g m

18560 J = 2452 J/g m

m = 7.6 g

7 0
3 years ago
The Strength of an electromagnet can be increased by reducing the number of turns on the wire coil true or false
ad-work [718]

Answer:the answer is true

Explanation:because as u add more coil/wrap more coil around the nail the field  get stronger and stronger to attract more nails

Hope this helped

5 0
3 years ago
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