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3 years ago

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Two violinists are playing their "A" strings. Each is perfectly tuned at 440 Hz and under 245 N of tension. If one violinist tur

ns her peg to tighten her A string to 251 N of tension, what beat frequency will result

1 answer:

const2013 [10]3 years ago

7 0

Answer:

450.78Hz

Explanation:

The frequency of wave in string is directly proportional to its tension. This is mathematically expressed as

F∝T

F = kT where k is the constant of proportionality

From the formula, k = F/T

F1/T1 = F2/T2 = k

If each violin is perfectly tuned at 440 Hz and under 245 N of tension then F1 = 440Hz, T1 = 245N

To get the frequency if the string A is tighten to 251N of tension,

T2 = 251N, F2 =?

Substituting the given values into the equation above to get F2 we have;

440/245 = F2/251

Cross multiplying

245F2 = 440×251

245F2 = 110,440

F2 = 110,440/245

F2 = 450.78Hz

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A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One

Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

$\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\$

To get the required work done, we will divide the mass by the speed of one knot to have:

$w=\frac{7230}{0.51}\\w= 14,176.47Joules$

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

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3 years ago

A plane drops a hamper of medical supplies from a height of 5000 m during a practice run over the ocean. The plane’s horizontal

Marizza181 [45]

Answer:

346.70015 m/s

Explanation:

In the x axis speed is

$v_x=149\ m/s$

In the y axis

$v_y=\sqrt{2gh}\\\Rightarrow v_y=\sqrt{2\times 9.8\times 5000}$

The resultant velocity is given by

$v=\sqrt{v_x^2+v_y^2}\\\Rightarrow v=\sqrt{149^2+2\times 9.8\times 5000}\\\Rightarrow v=346.70015\ m/s$

The magnitude of the overall velocity of the hamper at the instant it strikes the surface of the ocean is 346.70015 m/s

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3 years ago

Eric has a mass of 70 . He is standing on a scale in an elevator that is accelerating downward at 1.7 . What is the approximate

pychu [463]

Answer:

B)

Explanation:

The value the scale shows is the reaction force to the normal force (they are equal by Newton's 3rd Law) that the scale exerts on Eric.

The forces on Eric are his weight (downward) and this normal force (upward), so we can write the net force over him as (also using Newton's 2nd Law):

$F=W-N=ma$

which means

$N=W-ma=mg-ma=m(g-a)$

and for our values this is:

$N=mg-ma=(70kg)(9.8m/s^2-1.7m/s^2)=567N$

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3 years ago

A boy is pulling a cart by a force of 100N. The frictional force experienced by the cart is 20N. The force causing the motion of

Serga [27]

Answer:

3 is right i guss look : 100N-20N=80N

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2 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago