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2 years ago

7

What frequency will you hear if a truck is driving toward you at 20 m/s sounding a horn of frequency 300 Hz. Assume the speed of

sound to be 340 m/s.

1 answer:

Damm [24]2 years ago

6 0

Answer:

=319Hz

Explanation:

This question is on sound waves

General formulae;

Frequency of observer= f of source*[ (v of sound+ v of observer)/(v of sound-v of source)]

where f is frequency and v is velocity

Given;

Velocity of sound = 340m/s

frequency of source = 300 Hz

Velocity of observer=0m/s

Velocity of source/truck= 20m/s

Substitute values in the formulae;

Frequency of observer= 300× {(340+0)/(340-20)}

=300×{ 340/320}

=300×340/320

=318.75

=319Hz

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Answer:

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2 years ago

The length of vector A is 93.8 meters and the length of Ay is 38.4 meters, then the length of A, must be

vaieri [72.5K]

Answer:

85.556metres

Explanation:

Using pythagorean theorem

C²=A²+B²

we have c as the hypotenuse vector A thus:

93.8²=A²+38.4²

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2 years ago

g a small smetal sphere, carrying a net charge is held stationarry. what is the speed are 0.4 m apart

weeeeeb [17]

Complete Question

A small metal sphere, carrying a net charge q1=−2μC, is held in a stationary position by insulating supports. A second small metal sphere, with a net charge of q2= -8μC and mass 1.50g, is projected toward q1. When the two spheres are 0.80m apart, q2 is moving toward q1 with speed 20ms−1. Assume that the two spheres can be treated as point charges. You can ignore the force of gravity.The speed of q2 when the spheres are 0.400m apart is.

Answer:

The value $v_2 = 4 \sqrt{10} \ m/s$

Explanation:

From the question we are told that

The charge on the first sphere is $q_1 = 2\mu C = 2*10^{-6} \ C$

The charge on the second sphere is $q_2 = 8 \mu C = 8*10^{-6} \ C$

The mass of the second charge is $m = 1.50 \ g = 1.50 *10^{-3} \ kg$

The distance apart is $d = 0.4 \ m$

The speed of the second sphere is $v_1 = 20 \ ms^{-1}$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.8 \ m$ is mathematically represented

$Q = KE + U$

Here KE is the kinetic energy which is mathematically represented as

$KE = \frac{1 }{2} m (v_1)^2$

substituting value

$KE = \frac{1 }{2} * ( 1.50 *10^{-3}) (20 )^2$

$KE = 0.3 \ J$

And U is the potential energy which is mathematically represented as

$U = \frac{k * q_1 * q_2 }{d }$

substituting values

$U = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.8 }$

$U = 0.18 \ J$

So

$Q = 0.3 + 0.18$

$Q = 0.48 \ J$

Generally the total energy possessed by when $q_2$ and $q_1$ are separated by $0.4 \ m$ is mathematically represented

$Q_f = KE_f + U_f$

Here $KE_f$ is the kinetic energy which is mathematically represented as

$KE_f = \frac{1 }{2} m (v_2^2$

substituting value

$KE_f = \frac{1 }{2} * ( 1.50 *10^{-3}) (v_2 )^2$

$KE_f = 7.50 *10^{ -4} (v_2 )^2$

And $U_f$ is the potential energy which is mathematically represented as

$U_f = \frac{k * q_1 * q_2 }{d }$

substituting values

$U_f = \frac{9*10^9 * 2*10^{-6} * 8*10^{-6} }{0.4 }$

$U_f = 0.36 \ J$

From the law of energy conservation

$Q = Q_f$

So

$0.48 = 0.36 +(7.50 *10^{-4} v_2^2)$

$v_2 = 4 \sqrt{10} \ m/s$

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2 years ago

A 6 kg weight is lifted off the ground to a height that gives it 70.56 j of gravitational potential energy. what is its height?

Marina CMI [18]

GPE= 70.56 J -------------------> GPE= mgh-------------> X= height
70.56 = 6(kg) * 9.8(m/s/s) * X
70.56 = 58.8X
70.56/58.8= 58.8X/58.8
X= 1.2
The height is 1.2 feet or meters (whatever unit you are using in this problem)

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2 years ago

Read 2 more answers

A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca

UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial energy to gravitational potencial energy, so:

$E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J$

then we have only gravitational potencial energy when the ball is at its maximun height.

$E_g=m*g*h$

because all the energy was transformed Eg=Ee

$h=\frac{0.081J}{9.8m/s^2*m}$

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

$h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m$

6 0

2 years ago