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2 years ago

• A flame always points upwards. Why do you think this is so?

Physics

1 answer:

valina [46]2 years ago

3 0

Answer:

A flame always point upwards because the flame's gas is hotter than the surrounding air and, like you said, a hot gas is always lighter or less dense than a cold gas.

Explanation:

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What is gravity?what is one solar day?<br>

mafiozo [28]

Answer:

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6 0

3 years ago

The AC voltage source is connected to an inductor and a resistor in series. If the frequency of the source is increased the curr

DaniilM [7]

Answer:

If the frequency of the source is increased the current in the circuit will decrease.

Explanation:

The current through the circuit is given as;

$I = \frac{V}{Z}$

Where;

V is the voltage in the AC circuit

Z is the impedance

$Z = \sqrt{R^2 + X_L^2}$

Where;

R is the resistance

$X_L$ is the inductive reactance

$X_L$ = ωL = 2πfL

where;

L is the inductance

f is the frequency of the source

Finally, the current in the circuit is given as;

$I = \frac{V}{\sqrt{R^2 + (2\pi fL)^2} }$

From the equation above, an increase in frequency (f) will cause a decrease in current (I).

Therefore, If the frequency of the source is increased the current in the circuit will decrease.

5 0

3 years ago

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

7 0

3 years ago

PLEASE HELP ASAP WILL REWARD BRAINLIEST:))

laiz [17]

Answer:

120s^-1

Explanation:

v=12v

I=10A

and since rate is with time, therefore rate=energy/time.

H=IV

10×12=120/s

therefore the rate is 120s^-1

6 0

1 year ago

Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi

Marina86 [1]

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

$I = I_0 cos^2\theta$