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kvv77 [185]
3 years ago
15

HELP ASAP!!

Physics
1 answer:
andriy [413]3 years ago
3 0

Answer:

This is false becuase different object weigh different

Thank you!

Explanation:

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4) Write down the transformation of energy in torch light.<br>​
Elenna [48]

Answer:

<u>The transformation of energy in a torch light is as follows:</u>

1) When the torch is turned ON, the chemical energy in the batteries is converted into electrical energy.

2) The electrical energy is converted into heat and light energy. (We feel the torch to be hot after some time and we can see the light energy)

Hope this helped!

<h2>~AnonymousHelper1807</h2>
7 0
3 years ago
What is one joule work
natta225 [31]
It's a Newton Meter Those two are multiplied to get a joule.
8 0
3 years ago
A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe
Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

A_{t} = A_{0}\cdot e^{- \lambda t}    (1)

<em>where A_{t}: is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.</em>

To find A₀ we can use the following equation:  

A_{0} = N_{0} \lambda   (2)

<em>where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon </em>

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:        

N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}

<em>where m_{T}: is the tree's carbon mass, N_{A}: is the Avogadro's number and m_{^{12}C}: is the ¹²C mass.  </em>

N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C    

Similarly, from equation (2) λ is:

\lambda = \frac{Ln(2)}{t_{1/2}}

<em>where t 1/2: is the half-life of ¹⁴C= 5700 years </em>

\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}

So, the initial activity A₀ is:  

A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y    

Finally, we can calculate the time from equation (1):

t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y              

I hope it helps you!

4 0
3 years ago
The graph below shows the speed of a car as it drives along a racetrack.
nordsb [41]

Answer:

Average speed = distance/time

From 1 to 9 seconds:

Distance covered = 1 - 0.2 = 0.8 km

Time = 9 - 1 = 8 sec

Average speed = 0.8 km / 8 sec

Average speed = 0.1 km/s .

The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.

Surprise surprise ! The area under a speed/time graph is the distance covered during that time !

In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.

Explanation:

4 0
3 years ago
The amount of heat required to melt lead at it melting temperature is 23.0 kJ/kg. The amount of heat required to melt mercury at
svet-max [94.6K]

The addition of 24 kJ of energy will allow all of the mercury and lead to change from solid to liquid. The temperature of each substance will also increase.

4 0
3 years ago
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