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Harlamova29_29 [7]
3 years ago
14

Sound is an example of ?

Physics
2 answers:
wlad13 [49]3 years ago
8 0
Aaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
valentina_108 [34]3 years ago
3 0
A. Longitudinal wave
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. Suppose the mass of a fully loaded module in which astronauts take off from the Moon is 1.00 × 104 kg. The thrust of its engin
Viktor [21]

Answer:

a) The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

b) Then we can say that a thrust of 3*10^{4} N won't be able to lift off the module from the Earth because it's smaller than the module's weight (9.8 *10^{4} N).

Explanation:

a) During a vertical takeoff, the sum of the forces in the vertical axis will be equal to mass times the module's acceleration. In this this case, the thrust of the module's engines and the total module's weight are the only vertical forces. (In the Moon, the module's weight will be equal to its mass times the Moon's gravity acceleration)

T-(m*g)=m*a

Where:

T= thrust =3 *10^{4} N

m= module's mass =1 *10^{4} N

g= moon's gravity acceleration =1.623 \frac{m}{s^2}

a= module's acceleration during takeoff

Then, we can find the acceleration like this:

a=\frac{T}{m} -g=\frac{3*{10}^4 N}{1*{10}^4 kg}-1.623\frac{m}{s^2}

a=1.377 \frac{m}{s^2}

The module's acceleration in a vertical takeoff from the Moon will be 1.377 \frac{m}{s^2}

b) To takeoff, the module's engines must generate a thrust bigger than the module's weight, which will be its mass times the Earth's gravity acceleration.

weight=m*g=(1*{10}^4 kg)*(9.8 \frac{m}{s^2})=9.8 *10^{4} N

Then we can say that a thrust of 3*10^{4} N won't be able to lift off the module from the Earth because it's smaller than the module's weight (9.8 *10^{4} N).

8 0
4 years ago
As an intern at an engineering firm, you are asked to measure the moment of inertia of a large wheel for rotation about an axis
klio [65]

Hi there!

We can begin by finding the acceleration of the block.

Use the kinematic equation:

d = v_0t + \frac{1}{2}at^2

The block starts from rest, so:

d = \frac{1}{2}at^2\\\\12 = \frac{1}{2}a(4^2)\\\\\frac{24}{16} = a = 1.5 m/s^2

Now, we can do a summation of forces of the block using Newton's Second Law:

F = ma = m_bg - T

mb = mass of the block

T = tension of string

Solve for tension:

T = m_bg - ma = 8.2(9.8) - 8.2(1.5) = 68.06 N

Now, we can do a summation of torques for the wheel:

\Sigma \tau = rF\\\\\Sigma\tau = rT

Rewrite:

I\alpha = rT

We solved that the linear acceleration is 1.5 m/s², so we can solve for the angular acceleration using the following:

\alpha = a/r\\\\\alpha = 1.5/.42= 3.57 rad/sec^2

Now, plug in the values into the equation:

I(3.57) = (0.42)(68.06)\\\\I = (0.42)(68.06)/(3.57) = \boxed{8.00 kgm^2}

8 0
3 years ago
What to do with a brand new mobile phone battery
Rashid [163]
Place it in the back of your phone where the old battery was

7 0
4 years ago
Which one of the following is an example of an acid-base reaction?
Lina20 [59]

Answer:

A. 2HCl + Mg(OH)2 + 2H2O + MgCl2

hope it helps

6 0
3 years ago
A 41-turn square coil of area 0.074 m2 and a 123-turn circular coil are both placed perpendicular to the same changing magnetic
vesna_86 [32]

Answer:

<h3>The area of second coil is ≅ 0.025 m^{2}</h3>

Explanation:

Given :

No. of turns in the first coil N_{1} = 41

No. of turns in the second coil N_{2}  = 123

Area of first coil A_{1} = 0.074 m^{2}

According to the law of electromagnetic induction,

Induced emf = -N \frac{d \phi}{dt}

Where \phi = magnetic flux.

Since given in question emf of both coil is same so we compare above equation.

    -\frac{N_{1} d\phi _{1}   }{dt_{1} }  = -\frac{N_{2} d\phi _{2}   }{dt_{2} }

   \frac{N_{1} A_{1}   dB_{1}  }{dt_{1} }  = \frac{N_{2} A_{2} dB_{2}     }{dt_{2} }

        A_{2} = \frac{N_{1} A_{1}  }{N _{2}  }

        A_{2} = \frac{41 \times 0.074 }{123  }

        A_{2} = 0.0246 = 0.025 m^{2}

Therefore, the area of second coil is ≅ 0.025 m^{2}

4 0
4 years ago
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