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3 years ago

Air resistance is not a form of friction. * true or false

Physics

2 answers:

Oksanka [162]3 years ago

6 0

Answer: false

Explanation: air resistance is a type of frictional force!

full explanation:

Air resistance force is the force of air pushing against a moving object. Air resistance (also called drag) is a type of frictional force. Like all frictional forces, the force of air resistance always opposes the motion of an object. Usually, the air resistance force is not very strong.

Please let me know if i'm wrong! :))

professor190 [17]3 years ago

4 0

Answer:

<u>False</u>.

Explanation:

The air resistance is a special type of frictional force that acts upon objects as they travel through the air. The force of air resistance is often observed to oppose the motion of an object. This force will frequently be neglected due to its negligible magnitude (and due to the fact that it is mathematically difficult to predict its value). It is most noticeable for objects that travel at high speeds (e.g., a skydiver or a downhill skier) or for objects with large surface areas.

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The mole is 6.02 x 10 23 particles. If a person masses out the correct molar mass in grams for a substance then she would have a

Leto [7]

Answer:

1 mole of H2O is 18 grams (2 g H + 16 g Oxygen)

36 / 18 = 2

So 2 moles = 2 * 6.02E23 = 12.04E23 = 1.204E24

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3 years ago

1. Impulse is the product of force and distance. True or false?

daser333 [38]

Answer:

Explanation:

1. Impulse, I = F.t

The statement impulse is the product of Force and distance is false.

2. F = m g

Force necessary to lift the object depends on the mass of the object.

statement 2 is false.

3. Joule is equal to Newton times meter.

Statement 3 is false.

4. Work done to lift an object is correct statement.

Statement 4 is true.

5. Kinetic energy of an object is due to motion.

Statement 5 is false.

6. Stopping distance is directly proportional to the square of velocity.

If velocity is doubled, stopping distance is quadrupled.

Statement 6 is false.

8 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

3 years ago

An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference

kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

$K=qV$ (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

$K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J$

The kinetic energy of the particle is also:

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

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}$

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

$|F_B|=qvBsin(\theta)$

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

$|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N$

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

$r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm$

the radius of the trajectory of the electron is 10.1 cm

6 0

3 years ago

Air rushing against an airplane is an example of _____friction.

slava [35]

The word "static" would be known to be friction as air rushing against an airplane

6 0

3 years ago