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Serjik [45]
3 years ago
5

What does a battery powered heater warms up from ?

Physics
1 answer:
Westkost [7]3 years ago
8 0
Battery based heaters use electric resistance heating, which uses a lot of current (electricity) to create the heat.
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two object gravitationally attract with a force of 20N.if the distance between the two object centers is doubled, then what is t
Usimov [2.4K]
Gravity decreases with the square of the distance, so the new force is (20)/(2*2) = 5N.
8 0
3 years ago
Differentiate the following functions with respect to x <br>xsin x​
gtnhenbr [62]

Answer:

Here is your answer

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6 0
3 years ago
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What is the total energy of a particle with a rest mass of 1 gram moving with half the speed of light? 1 eV = 1.6 x 10^-19 J. An
GREYUIT [131]

Answer:

6.5 x 10^32 eV

Explanation:

mass of particle, mo = 1 g = 0.001 kg

velocity of particle, v = half of velocity of light = c / 2

c = 3 x 10^8 m/s

Energy associated to the particle

E = γ mo c^2

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

E=\frac{m_{0}c^}2}{\sqrt{1-\frac{c^{2}}{4c^{2}}}}

E=\frac{2m_{0}c^}2}{\sqrt{3}}

E=\frac{2\times0.001\times9\times10^{16}}{1.732}

E=1.04\times10^{14}J

Convert Joule into eV

1 eV = 1.6 x 10^-19 J

So, E=\frac{1.04\times10^{14}}{1.6\times10^{-19}}=6.5\times10^{32}eV

4 0
3 years ago
Panel A shows a ball shortly after being thrown upward. Panel B shows the same ball in an instant on its way down. Suppose air r
Anna11 [10]

Answer:

ur mom

Explanation:

6 0
2 years ago
A sample of a gas in a rigid container has an initial pressure of 5 atm at a temperature of 254.5 k. The temperature is decrease
skelet666 [1.2K]

The gas is in a rigid container: this means that its volume remains constant. Therefore, we can use Gay-Lussac law, which states that for a gas at constant volume, the pressure is directly proportional to the temperature. The law can be written as follows:

\frac{P_1}{T_1} = \frac{P_2}{T_2}

Where P1=5 atm is the initial pressure, T1=254.5 K is the initial temperature, P2 is the new pressure and T2=101.8 K is the new temperature. Re-arranging the equation and using the data of the problem, we can find P2:

P_2 = T_2 \frac{P_1}{T_1}=(101.8 K) \frac{5 atm}{254.5 K}=2 atm

So, the new pressure is 2 atm.

7 0
3 years ago
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