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3 years ago

A car horn creates a 595 Hz tone

Physics

1 answer:

aliina [53]3 years ago

3 0

Answer:

529 Hz

Explanation:

595(343-20)/(343 + 20) = 529

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A vertical spring gun is used to launch balls into the air. If the spring is compressed by 4.9 cm, the ball of mass 5.5 g is lau

AleksandrR [38]

We know, by conservation of energy :

$\dfrac{kx^2}{2}=mgh$

Therefore,

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}$

Putting given values, we get :

$\dfrac{x_1^2}{x_2^2}=\dfrac{h_1}{h_2}\\\\\dfrac{4.9^2}{x_2^2}=\dfrac{50.2}{2\times 50.2}\\\\x_2^2=2\times 4.9^2\\\\x_2 = 4.9\times \sqrt{2}\\\\x_2=6.93\ cm$

Therefore, the spring be compressed to 6.93 cm to send the ball twice as high.

Hence, this is the required solution.

6 0

3 years ago

Which of the following statements is true?

likoan [24]

Answer:

d.Energy as heat transferred into an object is determined by the amount of work done on the object.

Explanation:

4 0

3 years ago

Read 2 more answers

A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The d

ValentinkaMS [17]

Answer:

Explanation:

Initial velocity , u = 30 m/s

final velocity , v = 10 m/s

time , t = 5 seconds

1. Acceleration = v - u / t

= 10 - 30 / 5

= -20 / 5

= <u><em>- 4 m/s</em></u>

8 0

3 years ago

Meter #1 can measure voltage to within 0.1 volts. Meter #2 can measure voltage to within 0.01 volts.

Ann [662]

Meter #2 is more precise.

There's no information here that tells us which meter is more accurate.

4 0

3 years ago

Three equal point charges, each with charge 1.05 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

kirza4 [7]

Answer:

The value is $U = 0.06 \ J$

Explanation:

From the question we are told that

The value of charge on each three point charge is

$q_1 = q_2 = q_3 =q= 1.05 \mu C = 1.05 *10^{-6} \ C$

The length of the sides of the equilateral triangle is $r = 0.500 \$

Generally the total potential energy is mathematically represented as

$U = k * [ \frac{q_1 * q_2}{r} + \frac{q_2 * q_3}{r} + \frac{q_3 * q_1}{r} ]$

=> $U = k * 3 * \frac{q^2}{r}$

Here k is coulomb constant with value $k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $U = 9*10^9 * 3 * \frac{(1.05 *10^{-6})^2}{0.5 }$

$U = 0.06 \ J$

6 0

3 years ago