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Kazeer [188]
3 years ago
6

I need help someone answer, thanks!

Physics
2 answers:
Sonja [21]3 years ago
5 0

Answer:

The third one.

Explanation:

fhfztdgjkhhi

alisha [4.7K]3 years ago
3 0

Answer:

3rd one

Explanation:

Because the drum vibrates in a way that produces <u>compression waves. </u> The other options provide other waves, or no waves at all.

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A truck on a straight road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 20.0 m/s. Then the truck trav
aniked [119]
<span>You can use the equation
V_xf = V_xi + a_x(t)

V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0 t

Thus, solve for t and get 10seconds and then take 5 seconds to break after 20 seconds of driving so for

a) 10 + 20 + 5 = 35 seconds

</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
 
Need to find xf, knowing xi = 0

Thus, use the formula
 x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
 x_f = 100m
 
so for the first 10 seconds the truck traveled 100ms At a speed of 20m/s

20m/s = xm/20s 20*20 = x
x = 400
 
thus we have 100+400 = 500m then it slows down from 500m to x_f
 
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5) x_f = 500 + 50
x_f = 550
 
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s

thus V_xavg = 16m/s</span>
8 0
3 years ago
At what height h above the ground does the projectile have a speed of 0.5v?
dedylja [7]

Answer:height above ground at which projectile have velocity

0.5v is (0.0375v^2)

Explanation:

Using Vf = Vi - gt

Where Vf is final velocity

Vi is initial velocity

g is the acceleration due to gravity

t is the time taken

So, 0.5v = v - gt

t = 0.05v

Therefore height h = vt - 0.5gt^2

Subtitute t

h = 0.05v^2 - 0.0125v^2

h = 0.0375v^2

3 0
3 years ago
A converging-diverging nozzle has a throat area of 10 cm2 and an exit area of 28.96 cm2 . A normal shock stands in the exit when
Svetllana [295]

The tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

The given parameters:

  • <em>Throat area of the nozzle, </em>A^*<em> = 10 cm² = 0.001 m²</em>
  • <em>The exit area of the nozzle, A = 28.96 cm² = 0.002896 m²</em>
  • <em>Air pressure at sea level = 101.325 kPa</em>

The ratio of the areas of the converging-diverging nozzle is calculated as follows;

= \frac{A}{A^*} \\\\= \frac{0.002896}{0.001} \\\\= 2.896

From supersonic isentropic table, at \frac{A}{A^*} = 2.896, we can determine the following;

M_e = 2.6 \ kg/s\\\\\frac{P_o}{P_e} = 19.954

The tank pressure is calculated as follows;

\frac{P_o}{P_e} = 19.954 \\\\P_e = \frac{P_o}{19.954} \\\\P_e = \frac{101.325 \ kPa}{19.954} \\\\P_e = 5.08 \ kPa

Thus, the tank pressure is 5.08 kPa and the mass flow rate is 2.6 kg/s.

Learn  more about converging-diverging nozzle design here: brainly.com/question/13889483

8 0
2 years ago
Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the
Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

F = G \frac{m_1m_2}{r^2}

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

F = m_2a

Therefore, we can see that

a(r) = G \frac{m_1}{r^2}

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}

a(R) is actually the constant acceleration at sea level

and

a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}

Therefore:

a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}

6 0
4 years ago
Read 2 more answers
A mass is oscillating up and down on a spring. In the above graph of
melomori [17]

Answer:

<em>Amplitude= 8 m</em>

Explanation:

<u>The Amplitude of a Wave</u>

Sinusoidal Function  refers to a mathematical curve with a smooth and periodic oscillation. Its name comes from the sine function and is characterized by the amplitude or the maximum displacement or distance moved by a point on a vibrating body measured from its equilibrium position.

To calculate the amplitude from a graph, we measure the maximum point and the minimum point the wave reaches. Then we subtract both values and divide the result by 2.

The shown wave in the figure has a maximum value of 8 m and a minimum value of -8 m. The distance from the maximum to the minimum is 8-(-8)= 16 m, thus the amplitude is 16/2= 8m.

Amplitude= 8 m

5 0
3 years ago
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