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Calculate the energy consumed per day.

Physics

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1. a) The commercial unit for measurement of electrical energy consumption is the kilowatt-hour (kWh).

1. b)

Step 1: Calculate the energy consumed per day.
Power P=1000 W=1 kWP = 1000 \text{ W} = 1 \text{ kW}
Time per day tday=5 hourst_{\text{day}} = 5 \text{ hours}
Energy per day Eday=P×tday=1 kW×5 hours=5 kWhE_{\text{day}} = P \times t_{\text{day}} = 1 \text{ kW} \times 5 \text{ hours} = 5 \text{ kWh}

Step 2: Calculate the total energy consumed for 30 days.
Total days =30 days= 30 \text{ days}
Total energy Etotal=Eday×30=5 kWh/day×30 days=150 kWhE_{\text{total}} = E_{\text{day}} \times 30 = 5 \text{ kWh/day} \times 30 \text{ days} = 150 \text{ kWh}

Step 3: Calculate the total cost.
Cost per kWh =50 frs= 50 \text{ frs}
Total cost =Etotal×Cost per kWh=150 kWh×50 frs/kWh=7500 frs= E_{\text{total}} \times \text{Cost per kWh} = 150 \text{ kWh} \times 50 \text{ frs/kWh} = 7500 \text{ frs}

The total cost to run the heater for 30 days is 7500 frs\boxed{\text{7500 frs}}.

1. c)

Step 1: Calculate the operating current of the heater.
Power P=1000 WP = 1000 \text{ W}
Voltage V=220 VV = 220 \text{ V}
Current I=PVI = \frac{P}{V}

I=1000 W220 V=4.545 AI = \frac{1000 \text{ W}}{220 \text{ V}} = 4.545 \text{ A}

Step 2: Determine a suitable fuse value.
A fuse should have a rating slightly higher than the operating current to allow for minor fluctuations but blow if there's a significant overload. Common fuse ratings are 1 A, 2 A, 3 A, 5 A, 10 A, 13 A.
Since the operating current is approximately 4.55 A4.55 \text{ A}, a 5 A5 \text{ A} fuse would be the most appropriate choice.

The fuse value that would be good for the heater is 5 A\boxed{\text{5 A}}.

2. a)
The forces acting on the bucket are:

  • Tension (T) acting upwards from the string.
  • Weight (W) of the sand acting downwards due to gravity.
      ^ T
      |
      |
   -------
  /       \
 |  Sand   |
  \       /
   -------
      |
      v W

2. b)

Step 1: Identify the given mass.
Mass of sand m=25 kgm = 25 \text{ kg}

Step 2: Calculate the weight.
Assume acceleration due to gravity g=9.8 m/s2g = 9.8 \text{ m/s}^2.
Weight W=mgW = mg

W=25 kg×9.8 m/s2=245 NW = 25 \text{ kg} \times 9.8 \text{ m/s}^2 = 245 \text{ N}

The weight of the sand in the bucket is 245 N\boxed{\text{245 N}}.

2. c)
Mass is a measure of the amount of matter in an object and is constant regardless of location. Weight is the force of gravity acting on an object's mass and varies depending on the gravitational field strength.

2. d)

Step 1: State the condition for equilibrium.
If the system is in equilibrium, the net force acting on the bucket is zero. This means the upward force equals the downward force.
Upward force = Tension in the string (TT)
Downward force = Weight of the sand (WW)

Step 2: Calculate the support provided by the string.
Since the weight of the bucket is neglected, the support provided by the string is equal to the weight of the sand.
T=WT = W

T=245 NT = 245 \text{ N}

The support provided by the string is 245 N\boxed{\text{245 N}}.

3. a)

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