The cost of boiling 500cm3 of water using the 3kW kettle is 1.35 P.
<h3>
Cost of electricity for 3 kW kettle</h3>
The cost is calculated as follows;
1 unit = 9p /kWh
Total energy consumed by 3 kW kettle, E = P x t
where;
- P is power (kW)
- t is time in (hr)
E = 3 kW x (3 mins/60 mins/hr)
E = 0.15 kWh
Energy cost = 9 p/kWh x 0.15 kWh = 1.35 P
Thus, the cost of boiling 500cm3 of water using the 3kW kettle is 1.35 P.
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Answer:
if you slide a hockey puck on ice, it will eventually stop, because of friction on the ice
kite when the wind changes can be described by the first law
Explanation:
if you slide a hockey puck on ice, it will eventually stop, because of friction on the ice
kite when the wind changes can be described by the first law
F=ma
a=F/m
=825N/75kg
=825kg*m/75kg*s^2
=11m/s^2 in the direction of the force (ans)
<span>(1) </span>Through the Second
Law of motion, the equation for Force is:
F = m x a
Where
m is mass and a is acceleration (deceleration)
<span>(2) </span>Distance is
calculated through the equation,
D
= Vi^2 / 2a
Where
Vi is initial velocity
<span>(3) </span>Work is calculated
through the equation,
W = F x D
Substituting
the known values,
Part
A:
<span>(1) </span> F = (85
kg)(2 m/s^2) = 170 N
<span>(2) </span> D = (37
m/s)^2 / (2)(2 m/s^2) = 9.25 m
<span>(3) </span> W = (170
N)(9.25 m) = 1572.5 J
Part
B:
<span>(1) </span> F = (85 kg)(4
m/s^2) = 340 N
<span>(2) </span>D = (37 m/s)^2 /
(2)(4 m/s^2) = 4.625 m
<span>(3) </span><span> W = (340
N)(4.625 m) = 1572.5 J</span>
Answer:
Explanation:
Acceleration of particle A is 7.3 times the acceleration of particle B.
Let the acceleration of particle B is a, then the acceleration of particle A is
7.3 a.
Let the period of particle A is T and the period of particle B is 2.5 T.
Let the radius of particle A is RA and the radius of particle B is RB.
Use the formula for the centripetal force

So, 
The ratio of radius of A to the radius of B is given by


RA : RB = 1.17