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3 years ago

Hey, I need help can someone help me out, please?

Physics

2 answers:

wariber [46]3 years ago

8 0

Answer:

6 newton

7) f =ma = 15*15 = 225N

8) a= 100/20 = 5ms^-2

Explanation:

this is right pls mark as brainliest

yan [13]3 years ago

6 0

Explanation:

6) newton

7) f =ma = 15*15 = 225N

8) a= 100/20 = 5ms^-2

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What will be the fundamental frequency and first three overtones for a 26cm long organ at 20°C if it is open ( at 20°C, the spee

kakasveta [241]

Explanation:

The frequency of an organ pipe if it is open is given by :

$f=\dfrac{nv}{2l}$

v is speed of sound in air is 343 m/s at 20°C

For fundamental frequency, n = 1

$f=\dfrac{1\times 343}{2\times 0.26}\\\\f=659.61\ Hz$

First overtone frequency,

$f_1=2f\\\\f_1=2\times 659.61\\\\f_1=1319.22\ Hz$

Second overtone frequency,

$f_2=3f\\\\f_2=3\times 659.61\\\\f_2=1978.83\ Hz$

Third overtone frequency

$f_3=4f\\\\f_3=4\times 659.61\\\\f_3=2638.44\ Hz$

Hence, this is the required solution.

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3 years ago

Use the work energy theorem to solve each of these problems and neglect air resistance in all cases. a) A branch falls from the

Sav [38]

Answer:

a)43.8 m/s

b)103.54 m/s

Explanation:

Work energy theorem

Δ $E_{k}=$ Δ $E_{g}$

$E_{k2} - E_{k1}=-(E_{g2} - E_{g1})\\\\ \frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2} - mgh_{1} = 0$

a) In this part v1=0 and h2=0

$\frac{1}{2} mv^{2}_{2}- \frac{1}{2} mv^{2}_{1}+ mgh_{2}- mgh_{1} = 0\\\\\\\\\v_{2} =\sqrt{2gh_{1}} = \sqrt{2*9.8*98} =43.8 m/s$ =0

b) in this part v2=0, h1= 0, and h2= 545m

$\frac{1}{2} v^{2}_{1}= gh_{2} \\v_{1} =\sqrt{2gh_{2}}=103.54m/s$

5 0

3 years ago

Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of

Tresset [83]

Answer: Well they could go down a hill to gain more kinetic energy, or the answer can just be B. He can pedal harder to increase the rate to 10 meters/second. I hope I helped you.

8 0

3 years ago

Read 2 more answers

A girl launches a toy rocket from the ground. The engine experiences an average thrust of 5.26 N. The mass of the engine plus fu

Triss [41]

Answer: The answer for A is - v = 786.93 m/s

The answer for B is - v = 122.40 m/s

Explanation:

a) To find the average exhaust speed (v) of the engine we can use the following equation:

F = vΔm

Where:

F: is the thrust by the engine = 5.26 N

Δm: is the mass of the fuel = 12.7 g

Δt: is the time of the burning of fuel = 1.90 s

v = F×ΔT/ΔT

b) To calculate the final velocity of the rocket we need to find the acceleration.

The acceleration (a) can be calculated as follows:

a = F/M

In the above equation, m is an average between the mass of the engine plus the rocket case mass and the mass of the engine plus the rocket case minus the fuel mass:

m = (m_{engine} + m_{rocket}) + (m_{engine} + m_{rocket} - m_{fuel})}{2} = {2*m_{engine} + 2*m_{rocket} - m_{fuel}}{2} = 2*25.0 g + 2*63.0 g - 12.7 g}{2} = 81.65 g

Now, the acceleration is:

a = 5.26 N/81.65-t 10^³kg} = 64.42 m*s^²

Finally, the final velocity of the rocket can be calculated using the following kinematic equation:

v= v_{0} + at = 0 + 64.42 m*s^{-2}*1.90 s = 122.40 m/s

7 0

3 years ago

dos contenedores de objetos tienen 30 y 40 kg de masa respectivamente y están sobre una superficie horizontal sin rozamiento una

Dmitry_Shevchenko [17]

Answer:

The acceleration of each mass is the same, and approximately equal to $1.43\,\,\frac{m}{s^2}$

Explanation:

Notice that the mass of the whole system is: 30 kg + 40 kg = 70 kg

If we use a net force of 100 N on the combined system, we can find the acceleration imparted to the masses via Newton's second law:

$F_{net}= M\,*\,a\\100\,\,N = 70\,\,kg * a\\a = \frac{100}{70} \,\frac{m}{s^2} \\a\approx 1.43\,\frac{m}{s^2}$

4 0

3 years ago