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Blababa [14]
3 years ago
7

Does air resistence decrease with speed

Physics
2 answers:
Butoxors [25]3 years ago
8 0

Answer:

Yes, air resistance decrease with speed

Explanation:

Air resistance is a kind of fluid friction that acts when objects flow through fluids. It is affected by the velocity of moving objects and the area of the objects. When an object moves with a greater velocity the air resistance acting on them will be high, so the speed decreases.

Fluid friction acts in fluids namely liquids and gases. In liquids the friction is called buoyancy and in gases it is called air resistance or drag. When two objects having the same mass but different area move through air, the object with larger area will have less velocity compared with the other object.

dybincka [34]3 years ago
3 0

Answer:

yes

Explanation:

because when you slow down, the resistance slows with the speed.

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vlabodo [156]
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
6 0
3 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
3 years ago
Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth
USPshnik [31]

The time needed for the hammer to reach the surface of the Earth is 3.54 s.

<h3>Time of motion of the hammer</h3>

The time of motion is calculated as follows;

t = √(2h/g)

where;

  • h is height of fall
  • g is acceleration due to gravity

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

8 0
2 years ago
The amount of gravitational potential energy released as:_________.
nordsb [41]

Answer:

b.it depends on the distance it falls

8 0
3 years ago
A street light is mounted at the top of a 15-ft-tall pole. A man 6 ft tall walks away from the pole with a speed of 4 ft/s along
babymother [125]

Answer:

Explanation:

height of pole = 15 ft

height of man = 6 ft

Let the length of shadow is y .

According to the diagram

Let at any time the distance of man is x.

The two triangles are similar

\frac{y-x}{y}=\frac{6}{15}

15 y - 15 x = 6 y

9 y = 15 x

y=\frac{5}{3}x

Differentiate with respect to time.

\frac{dy}{dt}=\frac{5}{3}\frac{dx}{dt}

As given, dx/dt = 4 ft/s

\frac{dy}{dt}=\frac{5}{3}\times 4

\frac{dy}{dt}=\frac{20}{3} ft/s

6 0
3 years ago
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