All categories

3 years ago

17. What is movig from the sound source to the

Physics

2 answers:

Julli [10]3 years ago

7 0

Answer:

<h2>The answer is (Particles)</h2>

Explanation:

Crank3 years ago

7 0

Wind because it moves the fastest and it is really good source of energy

You might be interested in

Which statement correctly explains molecular motion in different states of matter using the kinetic theory?

Gelneren [198K]

the answer is the second one

8 0

3 years ago

When a second student joins the first, the height difference between the liquid levels in the right and left pistons is 40 cm .

vlada-n [284]

Answer:

m = 56.5 kg

Explanation:

Since the addition of mass on one piston caused a change in pressure head at the other. Diameter of the piston calculated is used as 0.46 m

Δm*g / Area = p * g * Δh ..... Eq1

$m = \frac{p*h*A}{g}$

$m = \frac{850 * 0.4 * pi*0.46^2}{4*9.81}\\\\m= 56.5 kg$

8 0

3 years ago

A car moves in a circular motion and it is subject to a centripetal acceleration of 24 m/s2. If the radius of the circular path

Softa [21]

Answer:

Explanation:

Centripetal acceleration is given by:

$a_{c} = v^{2}/r$

Thus, centripetal acceleration is inversely proportional to the radius. Thus, when radius will double, the centripetal acceleration will be halved.

3 0

3 years ago

The frequency of the George Washington Bridge is 2.05 Hz. What is its period?

Zina [86]

Answer: 0.5 seconds

Explanation:

Given that:

Frequency of the George Washington Bridge F = 2.05 Hz

Period T = ?

Recall that frequency is the number of cycles a wave can complete in one second. Hence, frequency is the inverse of period.

i.e F = 1/T

2.05Hz = 1/T

T = 1/2.05Hz

T = 0.488 seconds (Rounded to the nearest tenth as 0.5seconds)

Thus, the period of the George Washington Bridge is 0.5 seconds

6 0

3 years ago

How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s

poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

$W = \Delta KE$

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

Here,

m = mass

$v_{f,i}$ = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2$

$W_1 = 150(m) J$

Second case) When the particle goes from 20m/s to 30m/s

$\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2$

$\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2$

$W_1 = 250(m) J$

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

5 0

3 years ago