Answer:
54 Kobo
Explanation:
Units of <u>electricity</u> are measured in kilowatt hours (kWh).
Given information:
- 900 watt electric iron
- Appliance usage = 4 hours a week for 5 weeks
- Unit cost of electricity = 3 Kobo per kWh
<h3><u>Step 1</u></h3>
Convert the wattage of the electric iron from watts to kilowatts.
1000 watts (W) = 1 kilowatt (kW)
⇒ 900 watts = 1 ÷ 1000 = 0.9 kilowatts
This means that the power consumption of the electric iron is 0.9 kW per hour of use.
<h3><u>Step 2</u></h3>
Total hours spent pressing clothes:
= 4 hours per week for 5 weeks
= 4 × 5
= 20 hours
<u>Total power consumption</u>:
= number of kW × number of hours
= 0.9 × 20
= 18 kWh
<h3><u>Step 3</u></h3>
To find the <u>total cost</u>, multiply the total kWh by the cost per kWh:
⇒ Cost = 18 × 3 = 54 Kobo
Answer:
96w
Explanation:
p=Iv..where v=12 and I=8.0
3. Sand has a low specific heat compared to air
Answer:
Total impulse = 
= Initial momentum of the car
Explanation:
Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.
The final velocity of the car is 0 m/s as it is brought to rest.
Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.
The magnitude of impulse is the absolute value of the change in momentum.
Momentum of an object is equal to the product of its mass and velocity.
So, the initial momentum of the car is given as:
The final momentum of the car is given as:
Therefore, the impulse is given as:
Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
