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3 years ago

A beetle crawls from x=1.09m to x=2.55m in 12.8s. what was the average velocity?

Physics

1 answer:

Ivan3 years ago

6 0

The average velocity of the beetle is equal to the ratio between the displacement and the time:

$v=\frac{S}{t}$

The displacement of the beetle is

$S=2.55 m-1.09 m=1.46 m$

The time is t=12.8 s

Therefore, the average velocity is

$v=\frac{S}{t}=\frac{1.46 m}{12.8 s}=0.11 m/s$

in the positive x-direction.

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You blow across the open mouth of an empty test tube and producethe fundamental standing wave of the air column inside the testt

sertanlavr [38]

Answer:

(a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

Explanation:

Given that,

Length of tube = 11.0 cm

(a). We need to calculate the frequency of this standing wave

Using formula of fundamental frequency

$f_{1}=\dfrac{v}{4l}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times0.11}$

$f_{1}=781.81\ Hz$

$f_{1}=0.782\ kHz$

(b). If the test tube is half filled with water

When the tube is half filled the effective length of the tube is halved

We need to calculate the frequency

Using formula of fundamental frequency of the fundamental standing wave in the air

$f_{1}=\dfrac{v}{4(\dfrac{L}{2})}$

Put the value into the formula

$f_{1}=\dfrac{344}{4\times\dfrac{0.11}{2}}$

$f_{1}=1563.63\ Hz$

$f_{1}=1.563\ kHz$

Hence, (a). The frequency of this standing wave is 0.782 kHz.

(b). The frequency of the fundamental standing wave in the air is 1.563 kHz.

6 0

4 years ago

A body with mass of 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time tequals0 the bod

11Alexandr11 [23.1K]

Answer:

X(t) = 13/13 cos(12t+α)

C =13/13

π/6 s

Explanation:

(A) A body with mass 200 g is attached to the end of a spring that is stretched 20 cm by a force of 9 N. At time t = 0 the body is pulled 1 m in to the right, stretching the spring, 3 set in motion with an initial velocity of 5 m/s to the left.

(a) Find X(t) in the form c • cos(w_o*t— α)

(b) Find the amplitude 3 Period of motion of the body 1

mass: m = 200g = 0.200 kg

displacement: ΔX = 20 cm = 0.20 m

Spring Constant: K = 9/0.20 = 45 N/m

IV: X(0) = 1m V(0) = -5 m/s

Simple Harmonic Motion: c•cos(cosw_t— α) = X(t)

Circular Frequency: w_o = √k/m= √36/(0.20) = 13 rad/s

X(0) = 1m =c_1

X'(0) = V(0) = c_2*w_o/w_o

= -5/12 = c_2

"radians Technically Unitless"

Amplitude: c = √ci^2 + c^2 ==> √1^2 + (-5/12)^2 = 1 m =13/13 = c

X(t) = 13/13 cos(12t+α)

since, C>0 : damped forced vibration c_1>0, c_2>0

phase angle 2π+tan^-1(c_2/c_1)

=2π+tan^-1(-5/12/1)= 5.884

period: T =2π/w_o

=π/6 s

6 0

3 years ago

What materials have low thermal conductivity

777dan777 [17]

Aluminum<span> and </span><span>copper. Hope this helps.</span>

3 0

4 years ago

Read 2 more answers

If a 4 engine jet accelerates down a runway at 8.7 m/s^2, Suppose now that all 4 engins are operational on the jet from the prev

soldi70 [24.7K]

Answer:

$5.22m/s^2$

Explanation:

One of the first propulsion characteristics given in the example is that all engines are equal.

In this way if we have 4 engines running at the same time, it means that its capacity is 100%.

Under this premise, if 100% is found, the Jet is capable of reaching a speed of 8.7m / s ^ 2.

However, the question is, what would happen if 2.4 "Engines" now work.

To do this then we make a simple equivalence,

If 4 engines is the equivalent of 100%, when would it be 2.4 engines?

$X = \frac{2.4 * 100\%} { 4 }= 60\%$

In this way it would mean that the body could be driven to 60% of its total.

$Speed_{Decreased} = 8.7 * 60\% = 5.22 \frac{m}{s^2}$

3 0

3 years ago

A series RLC circuit is driven by a 1.0-kHz oscillator. The circuit parameters are Vms = 12 V, L = 5.0 mH, C= 4.0 uF, and R=102.

alexgriva [62]

Answer:

C) $12.3$ volts

Explanation:

f = frequency of oscillator = 1 kHz = 1000 Hz

$V_{rms}$ = 12 Volts

L = Inductance of Inductor = 5 mH = 0.005 H

$X_{L}$ = Inductive reactance

Inductive reactance is given as

$X_{L}$ = $2\pi fL$

$X_{L}$ = $2(3.14) (1000)(0.005)$

$X_{L} = 31.4$

$X_{C}$ = Capacitive reactance

Capacitive reactance is given as

$X_{C}=\frac{1}{2\pi fC}$

$X_{C}=\frac{1}{2(3.14)(1000)(4\times 10^{-6})}$

$X_{C} = 39.8$

Impedance of the circuit is given as

$z = \sqrt{R^{2}+(X_{L} - X_{C})^{2}}$

$z = \sqrt{102^{2}+(31.4 - 39.8)^{2}}$

$z = 102.35$

Rms Current flowing is given as

$i = \frac{V_{rms}}{z}$

$i = \frac{12}{102.35}$

$i = 0.12$ A

Rms potential difference across the resistor is given as

$V = i R$

$V = (0.12) (102)$

$V = 12.3$ volts

5 0

3 years ago