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2 years ago

Undersoil electrical heating systems are used in greenhouses. This system could also be

Physics

1 answer:

Nata [24]2 years ago

4 0

Answer:t

Explanation:

Given that,

Power output is

P = 0.50kW = 0.5 × 1000 = 500W.

We want to find energy transfer in

t = 2minutes = 2 × 60 = 120 seconds

Also, time to hour

t = 2/60 = 1/30 hour

Power is the rate of energy.

Therefore,

Power = energy / time taken

Energy = power × time taken

Energy = 0.5kW × 1/30 hr

Energy = 0.01667 kWh

Also, energy in Joule

Energy = power × time

Energy = 500W × 120s

Energy = 60,000J

Energy = 60kJ.

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One car has two and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their s

Leokris [45]

Answer:

$v_1 = 7.96 m/s$

$v_2 = 17.8 m/s$

Explanation:

Let the mass of the other car is "m" and its kinetic energy is

$K = \frac{1}{2}mv^2$

now the mass of the first car is two and half times and its kinetic energy is half that of other car

so we will have

$\frac{1}{2}(2.5m)v_1^2 = \frac{1}{2}(\frac{1}{2}mv^2)$

$2.5 v_1^2 = 0.5 v^2$

$v_1 = 0.447 v$

now speed of both cars is increased by value of 9 m/s

so now we will have same kinetic energy for both cars

$\frac{1}{2}(2.5 m)(0.447v + 9)^2 = \frac{1}{2}m(v + 9)^2$

$2.5(0.447 v + 9)^2 = (v + 9)^2$

$1.58(0.447v + 9) = v + 9$

$0.293v = 5.22$

$v = 17.8 m/s$

so speed of first car is

$v_1 = 0.447 v = 7.96 m/s$

$v_2 = 17.8 m/s$

3 0

2 years ago

: In heavy rushIn heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 minutes, then you have to stop for 3.5

zhuklara [117]

Guessing you want the average speed. We can multiple each speed by the time we spent going that speed, and them all together and then divide by the total time we spent in traffic to get the average speed. We spent a total of 7.5 minutes in traffic, so average speed = (12*1.5+0*3.5+15*2.5)/7.5 = 7.4 m/s

8 0

2 years ago

In the mobile m1=0.42 kg and m2=0.47 kg. What must the unknown distance to the nearest tenth of a cm be if the masses are to be

LuckyWell [14K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From he question we are told that

The first mass is $m_1 = 0.42kg$

The second mass is $m_2 = 0.47kg$

From the question we can see that at equilibrium the moment about the point where the string holding the bar (where $m_1 \ and \ m_2$ are hanged ) is attached is zero

Therefore we can say that

$m_1 * 15cm = m_2 * xcm$

Making x the subject of the formula

$x = \frac{m_1 * 15}{m_2}$

$= \frac{0.42 * 15}{0.47}$

$x = 13.4 cm$

Looking at the diagram we can see that the tension T on the string holding the bar where $m_1 \ and \ m_2$ are hanged is as a result of the masses ( $m_1 + m_2$ )