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2 years ago

Which of the following statements is FALSE?

Physics

2 answers:

forsale [732]2 years ago

6 0

Answer:

Explanation:

this is false as heat always moves form a warm place to a cool place not the other way around

m_a_m_a [10]2 years ago

5 0

D because heat doesn’t move to warm places it makes things or places warm

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If this atom has a balanced charged, how many protons would you expect to find in this atom?

avanturin [10]

The answer would be C because there is six electrons and so there will be six protons because the amount of protons and electrons have to be the same otherwise it would be an unbalanced particle and you wouldn't be able to touch the object it is in without worrying about something happening

3 0

3 years ago

Read 2 more answers

An electron that has an instantaneous velocity of ???? = 2.0 × 106 m ???? ???? + 3.0 × 106 m ???? ???? is moving through the uni

Setler79 [48]

Explanation:

It is given that,

Velocity of the electron, $v=(2\times 10^6i+3\times 10^6j)\ m/s$

Magnetic field, $B=(0.030i-0.15j)\ T$

Charge of electron, $q_e=-1.6\times 10^{-19}\ C$

(a) Let $F_e$ is the force on the electron due to the magnetic field. The magnetic force acting on it is given by :

$F_e=q_e(v\times B)$

$F_e=1.6\times 10^{-19}\times [(2\times 10^6i+3\times 10^6j)\times (0.030i-0.15j)]$

$F_e=-1.6\times 10^{-19}\times (-390000)(k)$

$F_e=6.24\times 10^{-14}k\ N$

(b) The charge of electron, $q_p=1.6\times 10^{-19}\ C$

The force acting on the proton is same as force on electron but in opposite direction i.e (-k). Hence, this is the required solution.

8 0

3 years ago

an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o

ZanzabumX [31]

Answer:

The maximum amplitude ( $V_{max}$ ) will be 7.96 V.

Explanation:

We know, for distortion free operation, the slew rate (S) of an OPAMP is written as

$S = 2 \pi f V_{max}$

where ' $f$ ' is the highest frequency signal.

Therefore, from the above equation we can write,

$&&5 \frac{V}{\mu s} = 2 \pi 100 kHz \times V_{max}\\&or,& V_{max} = \frac{5V}{10^{-6} s \times 2 \pi 100 \times 10^{3} Hz}\\&or,& V_{max} = \frac{5}{2 \pi \times 10^{-1}} V = 7.96 V$