Answer:
70m/s²
Explanation:
we will use the first equation of Dalton to find it
Answer:
E = 16.464 J
Explanation:
Given that,
Mass of tetherball, m = 0.8 kg
It is hit by a child and rises 2.1 m above the ground, h = 21. m
We need to find the maximum gravitational potential energy of the ball. The formula for the gravitational potential energy is given by :
E = mgh
g is acceleration due to gravity
E = 0.8 kg × 9.8 m/s² × 2.1 m
= 16.464 J
So, the maximum potential energy of the ball is 16.464 J.
I know i did part a correctly. heres what i did: momentum is conserved: m1 * u - m2 * u = m2 * v or (m1 - m2) * u = m2 * v Also, for an elastic head-on collision, we know that the relative velocity of approach = relative velocity of separation (from conservation of energy), or, for this problem, 2u = v Then (m1 - m2) * u = m2 * 2u m1 - m2 = 2 * m2 m1 = 3 * m2 m1 is the sphere that remained at rest (hence its absence from the RHS), so m2 = 0.3kg / 3 m2 = 0.1 kg b) this part confuses me, heres what i did (m1 - m2) * u = m2 * v (.3kg - .1kg)(2.0m/s) = .1kg * v .4 kg = .1 v v = 4 m/s What my teacher did: (.3g - .1g) * 2.0m/s = (.3g + .1g) * v I understand the left hand side but i dont get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero?? v = +1.00m/s since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean? thanks!
<span>Reference https://www.physicsforums.com/threads/elastic-collision-with-conservation-of-momentum-problem.651261...</span>
Answer:
$ 1.1
Explanation:
From the question given above, the following data were obtained:
Cost per kWh = $ 0.1
Current (I) = 10 A
Voltage (V) = 220 V
Time (t) = 5 h
Cost of operation =?
Next, we shall determine the power the electric oven. This can be obtained as follow:
Current (I) = 10 A
Voltage = 220 V
Power (P) =?
P = IV
P = 10 × 220
P = 2200 W
Next, we shall convert 2200 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
2200 W = 2200 W × 1 KW / 1000 W
2200 W = 2.2 KW
Thus, 2200 W is equivalent to 2.2 KW.
Next, we shall determine the energy consumed by the electric oven. This can be obtained as follow:
Power (P) = 2.2 KW
Time (t) = 5 h
Energy (E) =?
E = Pt
E = 2.2 × 5
E = 11 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
1 KWh cost $ 0.1
Therefore,
11 KWh will cost = 11 × 0.1
11 KWh will cost = $ 1.1
Therefore, the cost of operating the electric oven is $ 1.1
