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3 years ago

Which of these is an example of energy?

Physics

2 answers:

zysi [14]3 years ago

7 0

Answer:

The answer is Wind Blowing.

Explanation:

saul85 [17]3 years ago

4 0

Answer is A. Wind blowing

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____ are formed where bumps from two surfaces come into contact.

NemiM [27]

the answer is Friction.

hope this heslp

6 0

4 years ago

Read 2 more answers

A car is traveling at 20 km in 15 minutes. What is its average velocity

yarga [219]

The velocity is 4374.45 m/s.
I got the answer by using v=d/t.
20,000meters / 25min= 4,374.45 m/s.

7 0

2 years ago

A couch of mass 210 kg is being pushed across the floor with an applied force of 4100 N. The coefficient of kinetic friction bet

Dennis_Churaev [7]

Let us use the formula for Newton's Second Law of Motion:

Net force = Mass*Acceleration
Net force = Applied Force - μ*Normal Force
where μ is the coefficient of kinetic friction

Normal Force = Force due to gravity = mass*gravity
Normal Force = (210 kg)(9.81 m/s²) =<em> 2,060.1 N</em>

Then,
Net force = 4100 - 0.38*2060.1 = 3317.162 N

3317.162 N = (210 kg)(a)
Solving for acceleration,
<em>a = 15.796 m/s²</em>

8 0

4 years ago

A pipe that is open at both ends has a fundamental frequency of 320 Hz when the speed of sound in air is 331 m/s.

fenix001 [56]

Question

What is the length of the pipe?

Answer:

(a) 0.52m

(b) f2=640 Hz and f3=960 Hz

Explanation:

For an open pipe, the velocity is given by

$v=\frac {2Lf}{n}$

Making L the subject then

$L=\frac {nV}{2f}$

Where f is the frequency, L is the length, n is harmonic number, v is velocity

Substituting 1 for n, 320 Hz for f and 331 m/s for v then

$L=\frac {1*331}{2*320}=0.5171875\approx 0.52m$

(b)

The next two harmonics is given by

f2=2fi

f3=3fi

f2=3*320=640 Hz

f3=3*320=960 Hz

Alternatively, $f2=2\times \frac {v}{2L}$ and $f3=3\times \frac {v}{2L}$

$f2=2\times \frac {331}{2*0.52}=636.5 Hz\\f3=3\times \frac {331}{2*0.52}=954.8 Hz$

(c)

When v=367 m/s then

$f1= \frac {v}{2L}\\f1= \frac {367}{2*0.52}=352.9 Hz$

5 0

3 years ago

What is the electric flux ΦΦPhi through each of the six faces of the cube?

BlackZzzverrR [31]

Flux through each of the six faces of the cube: $\frac{q}{6\epsilon_0}$

Explanation:

In this problem, a charge $q$ is placed at the center of the cube.

We can apply Gauss Law, which states that the flux through the surface of the cube is equal to the charge contained within the cube divided by the vacuum permittivity; mathematically:

$\Phi_T=\frac{q}{\epsilon_0}$

where

$q$ is the charge

$\epsilon_0$ is the vacuum permittivity

Here we want to find the flux through each of the six faces of the cube.

By simmetry, we can say that the 6 faces are identical: therefore, the flux through each of them must be the same. This means that the flux through each faces is 1/6 of the total flux through the total surface, therefore:

$\Phi = \frac{1}{6}\Phi = \frac{q}{6\epsilon_0}$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0

3 years ago