All categories

3 years ago

A type of friction that occurs when air pushes against a moving object causing it to negatively accelerate

Physics

1 answer:

Romashka [77]3 years ago

6 0

Answer:

Air resistance

Answer B is correct

Explanation:

The friction that occurs when air pushes against a moving object causing it to negatively accelerate is called as air resistance.

hope this helps

brainliest appreciated

good luck! have a nice day!

You might be interested in

A car covers 120 km in 3 hours calculate its speed in m/s

8_murik_8 [283]

Answer:

11.1 m/s

Explanation:

120 km = 120 x 1,000 = 120,000 m

3 hours = 3 x 60 x 60 = 3600 x 3 = 10,800 s

speed = 120,000 / 10,800

= 1200/108

= 11.1 m/s

6 0

2 years ago

Read 2 more answers

Move the red arrow onto the picture to show how you think warm air would move over this area of land and water on a dark cold ni

Bess [88]

There’s no pictures?

7 0

2 years ago

Read 2 more answers

Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is

saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

$\dot{m_{i}} = \rho_{i} v_{i}A_{i}$

Where:

$v_{i}$ : is the initial velocity = 20 m/s

$A_{i}$ : is the inlet area of the nozzle = 60 cm²

$\rho_{i}$ : is the density of entrance = 2.21 kg/m³

$\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s$

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

$\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}$

$0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}$

$A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}$

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!

5 0

3 years ago

Read 2 more answers

If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f

dsp73

1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

where $\mu_0$ is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

2. The magnitude of the magnetic field halves

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

3. The force reverses direction

Explanation: the force exerted on a charged particle in a magnetic field is:

$F=qvB sin \theta$

where q is the charge, v is the speed of the particle, B is the magnetic field intensity and $\theta$ the angle between the direction of v and B. If the charge of the particle is switched from 2 µC to –2µC, the magnitude of the force does not change (because the absolute value of q does not change), however the charge q gets a negative sign (-), so the sign of the force changes and gets a negative sign too, so the force reverses direction.

7 0

3 years ago

Read 2 more answers

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

7 0

3 years ago